Is there any way to eliminate the singular point to solve this integral by hand or by approximations?
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$$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$
I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral?
calculus real-analysis analysis approximation
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up vote
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down vote
favorite
$$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$
I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral?
calculus real-analysis analysis approximation
1
What is $p$? Please give more information to make it complete.
– xbh
Sep 11 at 5:02
@xbh already fixed it, thanks for reminding.
– Mclalalala
Sep 11 at 5:10
Seems that $0$ is not singular here… [I might compute wrong]
– xbh
Sep 11 at 5:30
@xbh can you show me how to get the close formula for the integral above?
– Mclalalala
Sep 11 at 5:33
Sorry, i am working on it. It is highly impossible that this integral has a closed form.
– xbh
Sep 11 at 5:40
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$
I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral?
calculus real-analysis analysis approximation
$$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$
I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral?
calculus real-analysis analysis approximation
calculus real-analysis analysis approximation
edited Sep 11 at 5:07
asked Sep 11 at 4:48
Mclalalala
1268
1268
1
What is $p$? Please give more information to make it complete.
– xbh
Sep 11 at 5:02
@xbh already fixed it, thanks for reminding.
– Mclalalala
Sep 11 at 5:10
Seems that $0$ is not singular here… [I might compute wrong]
– xbh
Sep 11 at 5:30
@xbh can you show me how to get the close formula for the integral above?
– Mclalalala
Sep 11 at 5:33
Sorry, i am working on it. It is highly impossible that this integral has a closed form.
– xbh
Sep 11 at 5:40
|
show 3 more comments
1
What is $p$? Please give more information to make it complete.
– xbh
Sep 11 at 5:02
@xbh already fixed it, thanks for reminding.
– Mclalalala
Sep 11 at 5:10
Seems that $0$ is not singular here… [I might compute wrong]
– xbh
Sep 11 at 5:30
@xbh can you show me how to get the close formula for the integral above?
– Mclalalala
Sep 11 at 5:33
Sorry, i am working on it. It is highly impossible that this integral has a closed form.
– xbh
Sep 11 at 5:40
1
1
What is $p$? Please give more information to make it complete.
– xbh
Sep 11 at 5:02
What is $p$? Please give more information to make it complete.
– xbh
Sep 11 at 5:02
@xbh already fixed it, thanks for reminding.
– Mclalalala
Sep 11 at 5:10
@xbh already fixed it, thanks for reminding.
– Mclalalala
Sep 11 at 5:10
Seems that $0$ is not singular here… [I might compute wrong]
– xbh
Sep 11 at 5:30
Seems that $0$ is not singular here… [I might compute wrong]
– xbh
Sep 11 at 5:30
@xbh can you show me how to get the close formula for the integral above?
– Mclalalala
Sep 11 at 5:33
@xbh can you show me how to get the close formula for the integral above?
– Mclalalala
Sep 11 at 5:33
Sorry, i am working on it. It is highly impossible that this integral has a closed form.
– xbh
Sep 11 at 5:40
Sorry, i am working on it. It is highly impossible that this integral has a closed form.
– xbh
Sep 11 at 5:40
|
show 3 more comments
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1
What is $p$? Please give more information to make it complete.
– xbh
Sep 11 at 5:02
@xbh already fixed it, thanks for reminding.
– Mclalalala
Sep 11 at 5:10
Seems that $0$ is not singular here… [I might compute wrong]
– xbh
Sep 11 at 5:30
@xbh can you show me how to get the close formula for the integral above?
– Mclalalala
Sep 11 at 5:33
Sorry, i am working on it. It is highly impossible that this integral has a closed form.
– xbh
Sep 11 at 5:40