Interchange of a limit and integral
Clash Royale CLAN TAG#URR8PPP
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How would one show the following equality? Trying standard convergence theorems eg DCT (I can't find a choice of a dominating function which is also integrable) do not appear to be successful.
$$lim_a_n to 0^+ int_mathbbR e^frac sin(x)x dx=int_mathbbR frac sin(x)x dx $$
Thanks
integration limits
edited Aug 28 at 15:57
copper.hat
123k557156
asked Aug 28 at 15:30
kyhp
182
add a comment |Â
up vote
3
down vote
favorite
How would one show the following equality? Trying standard convergence theorems eg DCT (I can't find a choice of a dominating function which is also integrable) do not appear to be successful.
$$lim_a_n to 0^+ int_mathbbR e^frac sin(x)x dx=int_mathbbR frac sin(x)x dx $$
Thanks
integration limits
edited Aug 28 at 15:57
copper.hat
123k557156
asked Aug 28 at 15:30
kyhp
182
2
The function $x mapsto sin x over x$ is not integrable in the usual sense. It is an improper Lebesgue or Riemann integral.
– copper.hat
Aug 28 at 15:48
In other words, before we can try to prove the asserted equality, we need to have a precise definition of the integral on the right-hand side. What do you think it means? Is it, for example, a Cauchy principal value?
– Greg Martin
Aug 28 at 17:21
If I may recommend some general background to this, take a look at en.wikipedia.org/wiki/Sinc_function
– Mefitico
Aug 28 at 17:23
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How would one show the following equality? Trying standard convergence theorems eg DCT (I can't find a choice of a dominating function which is also integrable) do not appear to be successful.
$$lim_a_n to 0^+ int_mathbbR e^frac sin(x)x dx=int_mathbbR frac sin(x)x dx $$
Thanks
integration limits
edited Aug 28 at 15:57
copper.hat
123k557156
asked Aug 28 at 15:30
kyhp
182
How would one show the following equality? Trying standard convergence theorems eg DCT (I can't find a choice of a dominating function which is also integrable) do not appear to be successful.
$$lim_a_n to 0^+ int_mathbbR e^frac sin(x)x dx=int_mathbbR frac sin(x)x dx $$
Thanks
integration limits
edited Aug 28 at 15:57
copper.hat
123k557156
asked Aug 28 at 15:30
kyhp
182
edited Aug 28 at 15:57
copper.hat
123k557156
edited Aug 28 at 15:57
copper.hat
123k557156
edited Aug 28 at 15:57
copper.hat
123k557156
123k557156
asked Aug 28 at 15:30
kyhp
182
asked Aug 28 at 15:30
kyhp
182
asked Aug 28 at 15:30
kyhp
182
182
2
The function $x mapsto sin x over x$ is not integrable in the usual sense. It is an improper Lebesgue or Riemann integral.
– copper.hat
Aug 28 at 15:48
In other words, before we can try to prove the asserted equality, we need to have a precise definition of the integral on the right-hand side. What do you think it means? Is it, for example, a Cauchy principal value?
– Greg Martin
Aug 28 at 17:21
If I may recommend some general background to this, take a look at en.wikipedia.org/wiki/Sinc_function
– Mefitico
Aug 28 at 17:23
add a comment |Â
2
The function $x mapsto sin x over x$ is not integrable in the usual sense. It is an improper Lebesgue or Riemann integral.
– copper.hat
Aug 28 at 15:48
In other words, before we can try to prove the asserted equality, we need to have a precise definition of the integral on the right-hand side. What do you think it means? Is it, for example, a Cauchy principal value?
– Greg Martin
Aug 28 at 17:21
If I may recommend some general background to this, take a look at en.wikipedia.org/wiki/Sinc_function
– Mefitico
Aug 28 at 17:23
2
2
The function $x mapsto sin x over x$ is not integrable in the usual sense. It is an improper Lebesgue or Riemann integral.
– copper.hat
Aug 28 at 15:48
The function $x mapsto sin x over x$ is not integrable in the usual sense. It is an improper Lebesgue or Riemann integral.
– copper.hat
Aug 28 at 15:48
In other words, before we can try to prove the asserted equality, we need to have a precise definition of the integral on the right-hand side. What do you think it means? Is it, for example, a Cauchy principal value?
– Greg Martin
Aug 28 at 17:21
In other words, before we can try to prove the asserted equality, we need to have a precise definition of the integral on the right-hand side. What do you think it means? Is it, for example, a Cauchy principal value?
– Greg Martin
Aug 28 at 17:21
If I may recommend some general background to this, take a look at en.wikipedia.org/wiki/Sinc_function
– Mefitico
Aug 28 at 17:23
If I may recommend some general background to this, take a look at en.wikipedia.org/wiki/Sinc_function
– Mefitico
Aug 28 at 17:23
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
With $a > 0$ we have
$$int_mathbbRe^ fracsin xx , dx = 2 int_0^infty e^-ax fracsin xx , dx, $$
and it is enough to consider the limit of the integral on the RHS.
In that case, we want to show that
$$tag*lim_a to 0+int_0^infty e^-ax fracsin xx , dx = int_0^infty lim_a to 0+e^-ax fracsin xx , dx = int_0^infty fracsin xx , dx = fracpi2$$
We can switch the limit and the integral as long as the improper integral is unformly convergent for $a geqslant 0$. To prove this, note that for any $c > 0$
$$tag**left|int_0^infty e^-ax fracsin xx , dx - int_0^infty fracsin xx , dx right| \ leqslant left|int_0^c e^-ax fracsin xx , dx - int_0^c fracsin xx , dx right| + left|int_c^infty e^-ax fracsin xx , dx right| + left|int_c^infty fracsin xx , dx right| $$
Since the integrand is uniformly continuous for $(x,a) in [0,c] times [0,b]$ for any $b > 0$, there exists $delta > 0$ such that the first term on the RHS is less than $epsilon/3$ when $0 < a <delta$. This, of course, is the justification for switching the limit and integral over a bounded interval. By uniform convergence of the improper integral of $e^-ax sin x / x$ and the convergence of the improper integral of $sin x /x$, there exists $c > 0$ independent of $a$ such that second and third terms on the RHS are each less than $epsilon/3$ as well.
To confirm uniform convergence -- for any $c_2 > c_1 > 0$ we have by the second mean value theorem for integrals the existence of a point $xi$ with $c_1 < xi < c_2$ such that
$$left|int_c_1^c_2 e^-ax fracsin xx , dx right| = frace^-ac_1c_1left|int_c_1^xisin x , dx right| = frace^-ac_1c_1left|cos c_1 - cos xiright| leqslant frac2c_1$$
Now the RHS is smaller than $epsilon >0$ by choosing $c_1 > 2/ epsilon$ for any $a geqslant 0$ and, by the Cauchy criterion, the improper integral is uniformly convergent, Therefore (*) is true.
answered Aug 28 at 17:56
RRL
44.3k42362
I'm afraid I don't follow when you say we can switch the limit and the integral as long as the improper integral is unformly convergent. Isn't the integral just a numeric value as a function of a?
– kyhp
Aug 29 at 1:56
As you say the integral is of the form $F(a)= int_0^infty f(x,a) , dx$ and we are trying to show that $lim_a to 0+ F(a) = F(0) = int_0^infty f(x,0) , dx$. This is equivalent to switching the limit with the integral, since $lim_a to 0+ f(x,a) = f(x,0)$. To prove this we have to show that the LHS of (**) can be made smaller than any $epsilon >0$ when $0 < a < delta$ where $delta$ varies only with $epsilon$. It is a basic theorem that for an integral of $f(x,a)$ on a bounded interval where the integrand is continuous we can take the limit as $a to 0+$ inside the integral.
– RRL
Aug 29 at 3:05
The approach is to pick a $c > 0$ and show there is such a $delta$ such that if $0 < a < delta$, then the RHS of (**) is less than $epsilon$. This is accomplished by using the limit/integral switching result for the first term on the RHS when $c$ is fixed, after picking $c$ such that the second and third terms are small -- and to do this for the second term, $c$ can't vary with $a$ -- requiring uniform convergence of the improper integral $int_0^infty f(x,a) , dx$ for $a > 0$.
– RRL
Aug 29 at 3:11
More generally, this is a basic result proved in most real analysis books that $F(a) = int_0^infty f(x,a) , dx$ is continuous at $a_0$ when the integral converges uniformly and $f$ is continuous on $[0,infty) times I$ where $I$ is an interval containing $a_0$.
– RRL
Aug 29 at 3:14
Now I'm guessing you don't fully understand what it means for an improper integral to be uniformly convergent. You need to understand that for this to make sense. We say the improper integral $F(a) = int_0^infty f(x,a) , dx$ is uniformly convergent for $a geqslant 0$ if for every $epsilon > 0$ there exists $K(epsilon) > 0$ such that $left|int_0^c f(x,a) , dx - F(a)right| < epsilon$ for all $c > K(epsilon)$ and for all $a geqslant 0$. If $K$ depended on $a$, the integral could converge pointwise w/r $a$ but not uniformly.
– RRL
Aug 29 at 5:02
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
With $a > 0$ we have
$$int_mathbbRe^ fracsin xx , dx = 2 int_0^infty e^-ax fracsin xx , dx, $$
and it is enough to consider the limit of the integral on the RHS.
In that case, we want to show that
$$tag*lim_a to 0+int_0^infty e^-ax fracsin xx , dx = int_0^infty lim_a to 0+e^-ax fracsin xx , dx = int_0^infty fracsin xx , dx = fracpi2$$
We can switch the limit and the integral as long as the improper integral is unformly convergent for $a geqslant 0$. To prove this, note that for any $c > 0$
$$tag**left|int_0^infty e^-ax fracsin xx , dx - int_0^infty fracsin xx , dx right| \ leqslant left|int_0^c e^-ax fracsin xx , dx - int_0^c fracsin xx , dx right| + left|int_c^infty e^-ax fracsin xx , dx right| + left|int_c^infty fracsin xx , dx right| $$
Since the integrand is uniformly continuous for $(x,a) in [0,c] times [0,b]$ for any $b > 0$, there exists $delta > 0$ such that the first term on the RHS is less than $epsilon/3$ when $0 < a <delta$. This, of course, is the justification for switching the limit and integral over a bounded interval. By uniform convergence of the improper integral of $e^-ax sin x / x$ and the convergence of the improper integral of $sin x /x$, there exists $c > 0$ independent of $a$ such that second and third terms on the RHS are each less than $epsilon/3$ as well.
To confirm uniform convergence -- for any $c_2 > c_1 > 0$ we have by the second mean value theorem for integrals the existence of a point $xi$ with $c_1 < xi < c_2$ such that
$$left|int_c_1^c_2 e^-ax fracsin xx , dx right| = frace^-ac_1c_1left|int_c_1^xisin x , dx right| = frace^-ac_1c_1left|cos c_1 - cos xiright| leqslant frac2c_1$$
Now the RHS is smaller than $epsilon >0$ by choosing $c_1 > 2/ epsilon$ for any $a geqslant 0$ and, by the Cauchy criterion, the improper integral is uniformly convergent, Therefore (*) is true.
answered Aug 28 at 17:56
RRL
44.3k42362
I'm afraid I don't follow when you say we can switch the limit and the integral as long as the improper integral is unformly convergent. Isn't the integral just a numeric value as a function of a?
– kyhp
Aug 29 at 1:56
As you say the integral is of the form $F(a)= int_0^infty f(x,a) , dx$ and we are trying to show that $lim_a to 0+ F(a) = F(0) = int_0^infty f(x,0) , dx$. This is equivalent to switching the limit with the integral, since $lim_a to 0+ f(x,a) = f(x,0)$. To prove this we have to show that the LHS of (**) can be made smaller than any $epsilon >0$ when $0 < a < delta$ where $delta$ varies only with $epsilon$. It is a basic theorem that for an integral of $f(x,a)$ on a bounded interval where the integrand is continuous we can take the limit as $a to 0+$ inside the integral.
– RRL
Aug 29 at 3:05
The approach is to pick a $c > 0$ and show there is such a $delta$ such that if $0 < a < delta$, then the RHS of (**) is less than $epsilon$. This is accomplished by using the limit/integral switching result for the first term on the RHS when $c$ is fixed, after picking $c$ such that the second and third terms are small -- and to do this for the second term, $c$ can't vary with $a$ -- requiring uniform convergence of the improper integral $int_0^infty f(x,a) , dx$ for $a > 0$.
– RRL
Aug 29 at 3:11
More generally, this is a basic result proved in most real analysis books that $F(a) = int_0^infty f(x,a) , dx$ is continuous at $a_0$ when the integral converges uniformly and $f$ is continuous on $[0,infty) times I$ where $I$ is an interval containing $a_0$.
– RRL
Aug 29 at 3:14
Now I'm guessing you don't fully understand what it means for an improper integral to be uniformly convergent. You need to understand that for this to make sense. We say the improper integral $F(a) = int_0^infty f(x,a) , dx$ is uniformly convergent for $a geqslant 0$ if for every $epsilon > 0$ there exists $K(epsilon) > 0$ such that $left|int_0^c f(x,a) , dx - F(a)right| < epsilon$ for all $c > K(epsilon)$ and for all $a geqslant 0$. If $K$ depended on $a$, the integral could converge pointwise w/r $a$ but not uniformly.
– RRL
Aug 29 at 5:02
 |Â
show 1 more comment
up vote
2
down vote
accepted
With $a > 0$ we have
$$int_mathbbRe^ fracsin xx , dx = 2 int_0^infty e^-ax fracsin xx , dx, $$
and it is enough to consider the limit of the integral on the RHS.
In that case, we want to show that
$$tag*lim_a to 0+int_0^infty e^-ax fracsin xx , dx = int_0^infty lim_a to 0+e^-ax fracsin xx , dx = int_0^infty fracsin xx , dx = fracpi2$$
We can switch the limit and the integral as long as the improper integral is unformly convergent for $a geqslant 0$. To prove this, note that for any $c > 0$
$$tag**left|int_0^infty e^-ax fracsin xx , dx - int_0^infty fracsin xx , dx right| \ leqslant left|int_0^c e^-ax fracsin xx , dx - int_0^c fracsin xx , dx right| + left|int_c^infty e^-ax fracsin xx , dx right| + left|int_c^infty fracsin xx , dx right| $$
Since the integrand is uniformly continuous for $(x,a) in [0,c] times [0,b]$ for any $b > 0$, there exists $delta > 0$ such that the first term on the RHS is less than $epsilon/3$ when $0 < a <delta$. This, of course, is the justification for switching the limit and integral over a bounded interval. By uniform convergence of the improper integral of $e^-ax sin x / x$ and the convergence of the improper integral of $sin x /x$, there exists $c > 0$ independent of $a$ such that second and third terms on the RHS are each less than $epsilon/3$ as well.
To confirm uniform convergence -- for any $c_2 > c_1 > 0$ we have by the second mean value theorem for integrals the existence of a point $xi$ with $c_1 < xi < c_2$ such that
$$left|int_c_1^c_2 e^-ax fracsin xx , dx right| = frace^-ac_1c_1left|int_c_1^xisin x , dx right| = frace^-ac_1c_1left|cos c_1 - cos xiright| leqslant frac2c_1$$
Now the RHS is smaller than $epsilon >0$ by choosing $c_1 > 2/ epsilon$ for any $a geqslant 0$ and, by the Cauchy criterion, the improper integral is uniformly convergent, Therefore (*) is true.
answered Aug 28 at 17:56
RRL
44.3k42362
I'm afraid I don't follow when you say we can switch the limit and the integral as long as the improper integral is unformly convergent. Isn't the integral just a numeric value as a function of a?
– kyhp
Aug 29 at 1:56
As you say the integral is of the form $F(a)= int_0^infty f(x,a) , dx$ and we are trying to show that $lim_a to 0+ F(a) = F(0) = int_0^infty f(x,0) , dx$. This is equivalent to switching the limit with the integral, since $lim_a to 0+ f(x,a) = f(x,0)$. To prove this we have to show that the LHS of (**) can be made smaller than any $epsilon >0$ when $0 < a < delta$ where $delta$ varies only with $epsilon$. It is a basic theorem that for an integral of $f(x,a)$ on a bounded interval where the integrand is continuous we can take the limit as $a to 0+$ inside the integral.
– RRL
Aug 29 at 3:05
The approach is to pick a $c > 0$ and show there is such a $delta$ such that if $0 < a < delta$, then the RHS of (**) is less than $epsilon$. This is accomplished by using the limit/integral switching result for the first term on the RHS when $c$ is fixed, after picking $c$ such that the second and third terms are small -- and to do this for the second term, $c$ can't vary with $a$ -- requiring uniform convergence of the improper integral $int_0^infty f(x,a) , dx$ for $a > 0$.
– RRL
Aug 29 at 3:11
More generally, this is a basic result proved in most real analysis books that $F(a) = int_0^infty f(x,a) , dx$ is continuous at $a_0$ when the integral converges uniformly and $f$ is continuous on $[0,infty) times I$ where $I$ is an interval containing $a_0$.
– RRL
Aug 29 at 3:14
Now I'm guessing you don't fully understand what it means for an improper integral to be uniformly convergent. You need to understand that for this to make sense. We say the improper integral $F(a) = int_0^infty f(x,a) , dx$ is uniformly convergent for $a geqslant 0$ if for every $epsilon > 0$ there exists $K(epsilon) > 0$ such that $left|int_0^c f(x,a) , dx - F(a)right| < epsilon$ for all $c > K(epsilon)$ and for all $a geqslant 0$. If $K$ depended on $a$, the integral could converge pointwise w/r $a$ but not uniformly.
– RRL
Aug 29 at 5:02
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
With $a > 0$ we have
$$int_mathbbRe^ fracsin xx , dx = 2 int_0^infty e^-ax fracsin xx , dx, $$
and it is enough to consider the limit of the integral on the RHS.
In that case, we want to show that
$$tag*lim_a to 0+int_0^infty e^-ax fracsin xx , dx = int_0^infty lim_a to 0+e^-ax fracsin xx , dx = int_0^infty fracsin xx , dx = fracpi2$$
We can switch the limit and the integral as long as the improper integral is unformly convergent for $a geqslant 0$. To prove this, note that for any $c > 0$
$$tag**left|int_0^infty e^-ax fracsin xx , dx - int_0^infty fracsin xx , dx right| \ leqslant left|int_0^c e^-ax fracsin xx , dx - int_0^c fracsin xx , dx right| + left|int_c^infty e^-ax fracsin xx , dx right| + left|int_c^infty fracsin xx , dx right| $$
Since the integrand is uniformly continuous for $(x,a) in [0,c] times [0,b]$ for any $b > 0$, there exists $delta > 0$ such that the first term on the RHS is less than $epsilon/3$ when $0 < a <delta$. This, of course, is the justification for switching the limit and integral over a bounded interval. By uniform convergence of the improper integral of $e^-ax sin x / x$ and the convergence of the improper integral of $sin x /x$, there exists $c > 0$ independent of $a$ such that second and third terms on the RHS are each less than $epsilon/3$ as well.
To confirm uniform convergence -- for any $c_2 > c_1 > 0$ we have by the second mean value theorem for integrals the existence of a point $xi$ with $c_1 < xi < c_2$ such that
$$left|int_c_1^c_2 e^-ax fracsin xx , dx right| = frace^-ac_1c_1left|int_c_1^xisin x , dx right| = frace^-ac_1c_1left|cos c_1 - cos xiright| leqslant frac2c_1$$
Now the RHS is smaller than $epsilon >0$ by choosing $c_1 > 2/ epsilon$ for any $a geqslant 0$ and, by the Cauchy criterion, the improper integral is uniformly convergent, Therefore (*) is true.
answered Aug 28 at 17:56
RRL
44.3k42362
With $a > 0$ we have
$$int_mathbbRe^ fracsin xx , dx = 2 int_0^infty e^-ax fracsin xx , dx, $$
and it is enough to consider the limit of the integral on the RHS.
In that case, we want to show that
$$tag*lim_a to 0+int_0^infty e^-ax fracsin xx , dx = int_0^infty lim_a to 0+e^-ax fracsin xx , dx = int_0^infty fracsin xx , dx = fracpi2$$
We can switch the limit and the integral as long as the improper integral is unformly convergent for $a geqslant 0$. To prove this, note that for any $c > 0$
$$tag**left|int_0^infty e^-ax fracsin xx , dx - int_0^infty fracsin xx , dx right| \ leqslant left|int_0^c e^-ax fracsin xx , dx - int_0^c fracsin xx , dx right| + left|int_c^infty e^-ax fracsin xx , dx right| + left|int_c^infty fracsin xx , dx right| $$
Since the integrand is uniformly continuous for $(x,a) in [0,c] times [0,b]$ for any $b > 0$, there exists $delta > 0$ such that the first term on the RHS is less than $epsilon/3$ when $0 < a <delta$. This, of course, is the justification for switching the limit and integral over a bounded interval. By uniform convergence of the improper integral of $e^-ax sin x / x$ and the convergence of the improper integral of $sin x /x$, there exists $c > 0$ independent of $a$ such that second and third terms on the RHS are each less than $epsilon/3$ as well.
To confirm uniform convergence -- for any $c_2 > c_1 > 0$ we have by the second mean value theorem for integrals the existence of a point $xi$ with $c_1 < xi < c_2$ such that
$$left|int_c_1^c_2 e^-ax fracsin xx , dx right| = frace^-ac_1c_1left|int_c_1^xisin x , dx right| = frace^-ac_1c_1left|cos c_1 - cos xiright| leqslant frac2c_1$$
Now the RHS is smaller than $epsilon >0$ by choosing $c_1 > 2/ epsilon$ for any $a geqslant 0$ and, by the Cauchy criterion, the improper integral is uniformly convergent, Therefore (*) is true.
answered Aug 28 at 17:56
RRL
44.3k42362
edited Aug 29 at 3:02
answered Aug 28 at 17:56
RRL
44.3k42362
answered Aug 28 at 17:56
RRL
44.3k42362
answered Aug 28 at 17:56
RRL
44.3k42362
44.3k42362
I'm afraid I don't follow when you say we can switch the limit and the integral as long as the improper integral is unformly convergent. Isn't the integral just a numeric value as a function of a?
– kyhp
Aug 29 at 1:56
As you say the integral is of the form $F(a)= int_0^infty f(x,a) , dx$ and we are trying to show that $lim_a to 0+ F(a) = F(0) = int_0^infty f(x,0) , dx$. This is equivalent to switching the limit with the integral, since $lim_a to 0+ f(x,a) = f(x,0)$. To prove this we have to show that the LHS of (**) can be made smaller than any $epsilon >0$ when $0 < a < delta$ where $delta$ varies only with $epsilon$. It is a basic theorem that for an integral of $f(x,a)$ on a bounded interval where the integrand is continuous we can take the limit as $a to 0+$ inside the integral.
– RRL
Aug 29 at 3:05
The approach is to pick a $c > 0$ and show there is such a $delta$ such that if $0 < a < delta$, then the RHS of (**) is less than $epsilon$. This is accomplished by using the limit/integral switching result for the first term on the RHS when $c$ is fixed, after picking $c$ such that the second and third terms are small -- and to do this for the second term, $c$ can't vary with $a$ -- requiring uniform convergence of the improper integral $int_0^infty f(x,a) , dx$ for $a > 0$.
– RRL
Aug 29 at 3:11
More generally, this is a basic result proved in most real analysis books that $F(a) = int_0^infty f(x,a) , dx$ is continuous at $a_0$ when the integral converges uniformly and $f$ is continuous on $[0,infty) times I$ where $I$ is an interval containing $a_0$.
– RRL
Aug 29 at 3:14
Now I'm guessing you don't fully understand what it means for an improper integral to be uniformly convergent. You need to understand that for this to make sense. We say the improper integral $F(a) = int_0^infty f(x,a) , dx$ is uniformly convergent for $a geqslant 0$ if for every $epsilon > 0$ there exists $K(epsilon) > 0$ such that $left|int_0^c f(x,a) , dx - F(a)right| < epsilon$ for all $c > K(epsilon)$ and for all $a geqslant 0$. If $K$ depended on $a$, the integral could converge pointwise w/r $a$ but not uniformly.
– RRL
Aug 29 at 5:02
 |Â
show 1 more comment
I'm afraid I don't follow when you say we can switch the limit and the integral as long as the improper integral is unformly convergent. Isn't the integral just a numeric value as a function of a?
– kyhp
Aug 29 at 1:56
As you say the integral is of the form $F(a)= int_0^infty f(x,a) , dx$ and we are trying to show that $lim_a to 0+ F(a) = F(0) = int_0^infty f(x,0) , dx$. This is equivalent to switching the limit with the integral, since $lim_a to 0+ f(x,a) = f(x,0)$. To prove this we have to show that the LHS of (**) can be made smaller than any $epsilon >0$ when $0 < a < delta$ where $delta$ varies only with $epsilon$. It is a basic theorem that for an integral of $f(x,a)$ on a bounded interval where the integrand is continuous we can take the limit as $a to 0+$ inside the integral.
– RRL
Aug 29 at 3:05
The approach is to pick a $c > 0$ and show there is such a $delta$ such that if $0 < a < delta$, then the RHS of (**) is less than $epsilon$. This is accomplished by using the limit/integral switching result for the first term on the RHS when $c$ is fixed, after picking $c$ such that the second and third terms are small -- and to do this for the second term, $c$ can't vary with $a$ -- requiring uniform convergence of the improper integral $int_0^infty f(x,a) , dx$ for $a > 0$.
– RRL
Aug 29 at 3:11
More generally, this is a basic result proved in most real analysis books that $F(a) = int_0^infty f(x,a) , dx$ is continuous at $a_0$ when the integral converges uniformly and $f$ is continuous on $[0,infty) times I$ where $I$ is an interval containing $a_0$.
– RRL
Aug 29 at 3:14
Now I'm guessing you don't fully understand what it means for an improper integral to be uniformly convergent. You need to understand that for this to make sense. We say the improper integral $F(a) = int_0^infty f(x,a) , dx$ is uniformly convergent for $a geqslant 0$ if for every $epsilon > 0$ there exists $K(epsilon) > 0$ such that $left|int_0^c f(x,a) , dx - F(a)right| < epsilon$ for all $c > K(epsilon)$ and for all $a geqslant 0$. If $K$ depended on $a$, the integral could converge pointwise w/r $a$ but not uniformly.
– RRL
Aug 29 at 5:02
I'm afraid I don't follow when you say we can switch the limit and the integral as long as the improper integral is unformly convergent. Isn't the integral just a numeric value as a function of a?
– kyhp
Aug 29 at 1:56
I'm afraid I don't follow when you say we can switch the limit and the integral as long as the improper integral is unformly convergent. Isn't the integral just a numeric value as a function of a?
– kyhp
Aug 29 at 1:56
As you say the integral is of the form $F(a)= int_0^infty f(x,a) , dx$ and we are trying to show that $lim_a to 0+ F(a) = F(0) = int_0^infty f(x,0) , dx$. This is equivalent to switching the limit with the integral, since $lim_a to 0+ f(x,a) = f(x,0)$. To prove this we have to show that the LHS of (**) can be made smaller than any $epsilon >0$ when $0 < a < delta$ where $delta$ varies only with $epsilon$. It is a basic theorem that for an integral of $f(x,a)$ on a bounded interval where the integrand is continuous we can take the limit as $a to 0+$ inside the integral.
– RRL
Aug 29 at 3:05
As you say the integral is of the form $F(a)= int_0^infty f(x,a) , dx$ and we are trying to show that $lim_a to 0+ F(a) = F(0) = int_0^infty f(x,0) , dx$. This is equivalent to switching the limit with the integral, since $lim_a to 0+ f(x,a) = f(x,0)$. To prove this we have to show that the LHS of (**) can be made smaller than any $epsilon >0$ when $0 < a < delta$ where $delta$ varies only with $epsilon$. It is a basic theorem that for an integral of $f(x,a)$ on a bounded interval where the integrand is continuous we can take the limit as $a to 0+$ inside the integral.
– RRL
Aug 29 at 3:05
The approach is to pick a $c > 0$ and show there is such a $delta$ such that if $0 < a < delta$, then the RHS of (**) is less than $epsilon$. This is accomplished by using the limit/integral switching result for the first term on the RHS when $c$ is fixed, after picking $c$ such that the second and third terms are small -- and to do this for the second term, $c$ can't vary with $a$ -- requiring uniform convergence of the improper integral $int_0^infty f(x,a) , dx$ for $a > 0$.
– RRL
Aug 29 at 3:11
The approach is to pick a $c > 0$ and show there is such a $delta$ such that if $0 < a < delta$, then the RHS of (**) is less than $epsilon$. This is accomplished by using the limit/integral switching result for the first term on the RHS when $c$ is fixed, after picking $c$ such that the second and third terms are small -- and to do this for the second term, $c$ can't vary with $a$ -- requiring uniform convergence of the improper integral $int_0^infty f(x,a) , dx$ for $a > 0$.
– RRL
Aug 29 at 3:11
More generally, this is a basic result proved in most real analysis books that $F(a) = int_0^infty f(x,a) , dx$ is continuous at $a_0$ when the integral converges uniformly and $f$ is continuous on $[0,infty) times I$ where $I$ is an interval containing $a_0$.
– RRL
Aug 29 at 3:14
More generally, this is a basic result proved in most real analysis books that $F(a) = int_0^infty f(x,a) , dx$ is continuous at $a_0$ when the integral converges uniformly and $f$ is continuous on $[0,infty) times I$ where $I$ is an interval containing $a_0$.
– RRL
Aug 29 at 3:14
Now I'm guessing you don't fully understand what it means for an improper integral to be uniformly convergent. You need to understand that for this to make sense. We say the improper integral $F(a) = int_0^infty f(x,a) , dx$ is uniformly convergent for $a geqslant 0$ if for every $epsilon > 0$ there exists $K(epsilon) > 0$ such that $left|int_0^c f(x,a) , dx - F(a)right| < epsilon$ for all $c > K(epsilon)$ and for all $a geqslant 0$. If $K$ depended on $a$, the integral could converge pointwise w/r $a$ but not uniformly.
– RRL
Aug 29 at 5:02
Now I'm guessing you don't fully understand what it means for an improper integral to be uniformly convergent. You need to understand that for this to make sense. We say the improper integral $F(a) = int_0^infty f(x,a) , dx$ is uniformly convergent for $a geqslant 0$ if for every $epsilon > 0$ there exists $K(epsilon) > 0$ such that $left|int_0^c f(x,a) , dx - F(a)right| < epsilon$ for all $c > K(epsilon)$ and for all $a geqslant 0$. If $K$ depended on $a$, the integral could converge pointwise w/r $a$ but not uniformly.
– RRL
Aug 29 at 5:02
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2
The function $x mapsto sin x over x$ is not integrable in the usual sense. It is an improper Lebesgue or Riemann integral.
– copper.hat
Aug 28 at 15:48
In other words, before we can try to prove the asserted equality, we need to have a precise definition of the integral on the right-hand side. What do you think it means? Is it, for example, a Cauchy principal value?
– Greg Martin
Aug 28 at 17:21
If I may recommend some general background to this, take a look at en.wikipedia.org/wiki/Sinc_function
– Mefitico
Aug 28 at 17:23