Clarification needed for first proof of this question :$int_-infty^+infty f(x)dx = int_-infty^+infty fleft(x - frac1xright)dx?$
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I have tried many time to know why in the first given answer of this question used Bounds $-infty to infty$ in the second line however the variable change which has been used is clear dosn't give $-infty$ and $+infty$ , Probably a wrong typo of Author
integration proof-verification
asked Aug 8 at 16:37
zeraoulia rafik
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I have tried many time to know why in the first given answer of this question used Bounds $-infty to infty$ in the second line however the variable change which has been used is clear dosn't give $-infty$ and $+infty$ , Probably a wrong typo of Author
integration proof-verification
asked Aug 8 at 16:37
zeraoulia rafik
2,1451826
Focus on one of the terms: $int_-infty^0 f(x-1/x)dx$. Let $x=-e^theta$, which gives $theta=infty$ when $x=-infty$ and $theta=-infty$ when $x=0$. Everything seems fine?
– Alex R.
Aug 8 at 17:12
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up vote
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I have tried many time to know why in the first given answer of this question used Bounds $-infty to infty$ in the second line however the variable change which has been used is clear dosn't give $-infty$ and $+infty$ , Probably a wrong typo of Author
integration proof-verification
asked Aug 8 at 16:37
zeraoulia rafik
2,1451826
I have tried many time to know why in the first given answer of this question used Bounds $-infty to infty$ in the second line however the variable change which has been used is clear dosn't give $-infty$ and $+infty$ , Probably a wrong typo of Author
integration proof-verification
asked Aug 8 at 16:37
zeraoulia rafik
2,1451826
asked Aug 8 at 16:37
zeraoulia rafik
2,1451826
asked Aug 8 at 16:37
zeraoulia rafik
2,1451826
asked Aug 8 at 16:37
zeraoulia rafik
2,1451826
2,1451826
Focus on one of the terms: $int_-infty^0 f(x-1/x)dx$. Let $x=-e^theta$, which gives $theta=infty$ when $x=-infty$ and $theta=-infty$ when $x=0$. Everything seems fine?
– Alex R.
Aug 8 at 17:12
add a comment |Â
Focus on one of the terms: $int_-infty^0 f(x-1/x)dx$. Let $x=-e^theta$, which gives $theta=infty$ when $x=-infty$ and $theta=-infty$ when $x=0$. Everything seems fine?
– Alex R.
Aug 8 at 17:12
Focus on one of the terms: $int_-infty^0 f(x-1/x)dx$. Let $x=-e^theta$, which gives $theta=infty$ when $x=-infty$ and $theta=-infty$ when $x=0$. Everything seems fine?
– Alex R.
Aug 8 at 17:12
Focus on one of the terms: $int_-infty^0 f(x-1/x)dx$. Let $x=-e^theta$, which gives $theta=infty$ when $x=-infty$ and $theta=-infty$ when $x=0$. Everything seems fine?
– Alex R.
Aug 8 at 17:12
add a comment |Â
1 Answer
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Your question is not at all clear. But, I think you are asking.
When go from here:
$int_-infty^inftyfleft(x-x^-1right)dx=int_0^inftyfleft(x-x^-1right)dx+int_-infty^0fleft(x-x^-1right)dx$
to here:
$int_-infty^inftyf(2sinhtheta),e^thetadtheta+int_-infty^inftyf(2sinhtheta),e^-thetadtheta$
How do we justify the change to the limits of integration?
In the left hand integral above, we make the substitution $x = e^theta$
This substitution acts on the limits of integration. And we must find values of $theta$ that correspond to the $x$ values associated with $0, infty.$ Alternatively, we can invert the substitution. $theta = ln x, lim_limitsxto 0^+ ln x = -infty, lim_limitsxto infty ln x = infty.$
And for the right hand intgral we do something very similar, except our substituion is $x = e^-theta$
answered Aug 8 at 17:15
Doug M
39.3k31749
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your question is not at all clear. But, I think you are asking.
When go from here:
$int_-infty^inftyfleft(x-x^-1right)dx=int_0^inftyfleft(x-x^-1right)dx+int_-infty^0fleft(x-x^-1right)dx$
to here:
$int_-infty^inftyf(2sinhtheta),e^thetadtheta+int_-infty^inftyf(2sinhtheta),e^-thetadtheta$
How do we justify the change to the limits of integration?
In the left hand integral above, we make the substitution $x = e^theta$
This substitution acts on the limits of integration. And we must find values of $theta$ that correspond to the $x$ values associated with $0, infty.$ Alternatively, we can invert the substitution. $theta = ln x, lim_limitsxto 0^+ ln x = -infty, lim_limitsxto infty ln x = infty.$
And for the right hand intgral we do something very similar, except our substituion is $x = e^-theta$
answered Aug 8 at 17:15
Doug M
39.3k31749
add a comment |Â
up vote
0
down vote
Your question is not at all clear. But, I think you are asking.
When go from here:
$int_-infty^inftyfleft(x-x^-1right)dx=int_0^inftyfleft(x-x^-1right)dx+int_-infty^0fleft(x-x^-1right)dx$
to here:
$int_-infty^inftyf(2sinhtheta),e^thetadtheta+int_-infty^inftyf(2sinhtheta),e^-thetadtheta$
How do we justify the change to the limits of integration?
In the left hand integral above, we make the substitution $x = e^theta$
This substitution acts on the limits of integration. And we must find values of $theta$ that correspond to the $x$ values associated with $0, infty.$ Alternatively, we can invert the substitution. $theta = ln x, lim_limitsxto 0^+ ln x = -infty, lim_limitsxto infty ln x = infty.$
And for the right hand intgral we do something very similar, except our substituion is $x = e^-theta$
answered Aug 8 at 17:15
Doug M
39.3k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your question is not at all clear. But, I think you are asking.
When go from here:
$int_-infty^inftyfleft(x-x^-1right)dx=int_0^inftyfleft(x-x^-1right)dx+int_-infty^0fleft(x-x^-1right)dx$
to here:
$int_-infty^inftyf(2sinhtheta),e^thetadtheta+int_-infty^inftyf(2sinhtheta),e^-thetadtheta$
How do we justify the change to the limits of integration?
In the left hand integral above, we make the substitution $x = e^theta$
This substitution acts on the limits of integration. And we must find values of $theta$ that correspond to the $x$ values associated with $0, infty.$ Alternatively, we can invert the substitution. $theta = ln x, lim_limitsxto 0^+ ln x = -infty, lim_limitsxto infty ln x = infty.$
And for the right hand intgral we do something very similar, except our substituion is $x = e^-theta$
answered Aug 8 at 17:15
Doug M
39.3k31749
Your question is not at all clear. But, I think you are asking.
When go from here:
$int_-infty^inftyfleft(x-x^-1right)dx=int_0^inftyfleft(x-x^-1right)dx+int_-infty^0fleft(x-x^-1right)dx$
to here:
$int_-infty^inftyf(2sinhtheta),e^thetadtheta+int_-infty^inftyf(2sinhtheta),e^-thetadtheta$
How do we justify the change to the limits of integration?
In the left hand integral above, we make the substitution $x = e^theta$
This substitution acts on the limits of integration. And we must find values of $theta$ that correspond to the $x$ values associated with $0, infty.$ Alternatively, we can invert the substitution. $theta = ln x, lim_limitsxto 0^+ ln x = -infty, lim_limitsxto infty ln x = infty.$
And for the right hand intgral we do something very similar, except our substituion is $x = e^-theta$
answered Aug 8 at 17:15
Doug M
39.3k31749
answered Aug 8 at 17:15
Doug M
39.3k31749
answered Aug 8 at 17:15
Doug M
39.3k31749
answered Aug 8 at 17:15
Doug M
39.3k31749
39.3k31749
add a comment |Â
add a comment |Â
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Focus on one of the terms: $int_-infty^0 f(x-1/x)dx$. Let $x=-e^theta$, which gives $theta=infty$ when $x=-infty$ and $theta=-infty$ when $x=0$. Everything seems fine?
– Alex R.
Aug 8 at 17:12