How to calculate the integral $ I= int_0^{infty} frac{t^{a-1}}{t-x} dt$
$begingroup$
How to calculate the integral
$$ I= int_0^{infty} frac{t^{a-1}}{t-x} dt$$
using contour integration. ($x>0$ is fixed, $a in mathbb{R}$)
Approach: I calculated the values of $a$ for which the integral is convergent. The result was that $0<a<1$. We have two poles, at $t=0$ and $t=x$. I tried a circular key hole contour that excludes $t=0$ and $t=x$. The value of the integral integrated over the circular curves are zero I thougt. But the sum of the integrals integrated over the horizontal segments are $(1-e^{i(a-1)2pi})I$, so the integral is zero. But I made a mistake I guess. Can someone help me? Thanks in advance.
EDIT: I tried to use the same keyhole in this other post: Integration of $ln $ around a keyhole contour
complex-analysis contour-integration residue-calculus
asked Jun 22 '17 at 16:51
bobbob
686518
$endgroup$
add a comment |
$begingroup$
How to calculate the integral
$$ I= int_0^{infty} frac{t^{a-1}}{t-x} dt$$
using contour integration. ($x>0$ is fixed, $a in mathbb{R}$)
Approach: I calculated the values of $a$ for which the integral is convergent. The result was that $0<a<1$. We have two poles, at $t=0$ and $t=x$. I tried a circular key hole contour that excludes $t=0$ and $t=x$. The value of the integral integrated over the circular curves are zero I thougt. But the sum of the integrals integrated over the horizontal segments are $(1-e^{i(a-1)2pi})I$, so the integral is zero. But I made a mistake I guess. Can someone help me? Thanks in advance.
EDIT: I tried to use the same keyhole in this other post: Integration of $ln $ around a keyhole contour
complex-analysis contour-integration residue-calculus
asked Jun 22 '17 at 16:51
bobbob
686518
$endgroup$
$begingroup$
This integral must be computed in the Cauchy sense, otehrwise it is undefined because of the pole at $t=x$.
$endgroup$
– Yves Daoust
Jun 22 '17 at 16:59
$begingroup$
WA gives the result containing a hypergeometric function
$endgroup$
– Dr. Sonnhard Graubner
Jun 22 '17 at 17:01
$begingroup$
@YvesDaoust But the pole $t=x$ is not in the contour. Or what am I doing wrong?
$endgroup$
– bob
Jun 22 '17 at 17:05
$begingroup$
The pole is absolutely on the real axis. This integral must be computed in the Cauchy sense, otherwise it is undefined.
$endgroup$
– Yves Daoust
Jun 22 '17 at 17:24
$begingroup$
@Dr.SonnhardGraubner What on earth are you thinking? The answer can be found in closed form.
$endgroup$
– Mark Viola
Jun 23 '17 at 3:03
add a comment |
$begingroup$
How to calculate the integral
$$ I= int_0^{infty} frac{t^{a-1}}{t-x} dt$$
using contour integration. ($x>0$ is fixed, $a in mathbb{R}$)
Approach: I calculated the values of $a$ for which the integral is convergent. The result was that $0<a<1$. We have two poles, at $t=0$ and $t=x$. I tried a circular key hole contour that excludes $t=0$ and $t=x$. The value of the integral integrated over the circular curves are zero I thougt. But the sum of the integrals integrated over the horizontal segments are $(1-e^{i(a-1)2pi})I$, so the integral is zero. But I made a mistake I guess. Can someone help me? Thanks in advance.
EDIT: I tried to use the same keyhole in this other post: Integration of $ln $ around a keyhole contour
complex-analysis contour-integration residue-calculus
asked Jun 22 '17 at 16:51
bobbob
686518
$endgroup$
How to calculate the integral
$$ I= int_0^{infty} frac{t^{a-1}}{t-x} dt$$
using contour integration. ($x>0$ is fixed, $a in mathbb{R}$)
Approach: I calculated the values of $a$ for which the integral is convergent. The result was that $0<a<1$. We have two poles, at $t=0$ and $t=x$. I tried a circular key hole contour that excludes $t=0$ and $t=x$. The value of the integral integrated over the circular curves are zero I thougt. But the sum of the integrals integrated over the horizontal segments are $(1-e^{i(a-1)2pi})I$, so the integral is zero. But I made a mistake I guess. Can someone help me? Thanks in advance.
EDIT: I tried to use the same keyhole in this other post: Integration of $ln $ around a keyhole contour
complex-analysis contour-integration residue-calculus
complex-analysis contour-integration residue-calculus
asked Jun 22 '17 at 16:51
bobbob
686518
asked Jun 22 '17 at 16:51
bobbob
686518
asked Jun 22 '17 at 16:51
bobbob
686518
asked Jun 22 '17 at 16:51
bobbob
686518
asked Jun 22 '17 at 16:51
bobbob
686518
686518
$begingroup$
This integral must be computed in the Cauchy sense, otehrwise it is undefined because of the pole at $t=x$.
$endgroup$
– Yves Daoust
Jun 22 '17 at 16:59
$begingroup$
WA gives the result containing a hypergeometric function
$endgroup$
– Dr. Sonnhard Graubner
Jun 22 '17 at 17:01
$begingroup$
@YvesDaoust But the pole $t=x$ is not in the contour. Or what am I doing wrong?
$endgroup$
– bob
Jun 22 '17 at 17:05
$begingroup$
The pole is absolutely on the real axis. This integral must be computed in the Cauchy sense, otherwise it is undefined.
$endgroup$
– Yves Daoust
Jun 22 '17 at 17:24
$begingroup$
@Dr.SonnhardGraubner What on earth are you thinking? The answer can be found in closed form.
$endgroup$
– Mark Viola
Jun 23 '17 at 3:03
add a comment |
$begingroup$
This integral must be computed in the Cauchy sense, otehrwise it is undefined because of the pole at $t=x$.
$endgroup$
– Yves Daoust
Jun 22 '17 at 16:59
$begingroup$
WA gives the result containing a hypergeometric function
$endgroup$
– Dr. Sonnhard Graubner
Jun 22 '17 at 17:01
$begingroup$
@YvesDaoust But the pole $t=x$ is not in the contour. Or what am I doing wrong?
$endgroup$
– bob
Jun 22 '17 at 17:05
$begingroup$
The pole is absolutely on the real axis. This integral must be computed in the Cauchy sense, otherwise it is undefined.
$endgroup$
– Yves Daoust
Jun 22 '17 at 17:24
$begingroup$
@Dr.SonnhardGraubner What on earth are you thinking? The answer can be found in closed form.
$endgroup$
– Mark Viola
Jun 23 '17 at 3:03
$begingroup$
This integral must be computed in the Cauchy sense, otehrwise it is undefined because of the pole at $t=x$.
$endgroup$
– Yves Daoust
Jun 22 '17 at 16:59
$begingroup$
This integral must be computed in the Cauchy sense, otehrwise it is undefined because of the pole at $t=x$.
$endgroup$
– Yves Daoust
Jun 22 '17 at 16:59
$begingroup$
WA gives the result containing a hypergeometric function
$endgroup$
– Dr. Sonnhard Graubner
Jun 22 '17 at 17:01
$begingroup$
WA gives the result containing a hypergeometric function
$endgroup$
– Dr. Sonnhard Graubner
Jun 22 '17 at 17:01
$begingroup$
@YvesDaoust But the pole $t=x$ is not in the contour. Or what am I doing wrong?
$endgroup$
– bob
Jun 22 '17 at 17:05
$begingroup$
@YvesDaoust But the pole $t=x$ is not in the contour. Or what am I doing wrong?
$endgroup$
– bob
Jun 22 '17 at 17:05
$begingroup$
The pole is absolutely on the real axis. This integral must be computed in the Cauchy sense, otherwise it is undefined.
$endgroup$
– Yves Daoust
Jun 22 '17 at 17:24
$begingroup$
The pole is absolutely on the real axis. This integral must be computed in the Cauchy sense, otherwise it is undefined.
$endgroup$
– Yves Daoust
Jun 22 '17 at 17:24
$begingroup$
@Dr.SonnhardGraubner What on earth are you thinking? The answer can be found in closed form.
$endgroup$
– Mark Viola
Jun 23 '17 at 3:03
$begingroup$
@Dr.SonnhardGraubner What on earth are you thinking? The answer can be found in closed form.
$endgroup$
– Mark Viola
Jun 23 '17 at 3:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since the integrand is singular at $t=x$, this must be understood in the Cauchy principal value sense. Your contour should include little detours around $x=a$ as well as $0$.
The integrals over the arcs around $x$ are not $0$. For the top part, I get $-i pi x^{a-1}$; for the bottom part, $-i pi x^{a-1} e^{2 pi i (a-1)}$.
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
Oh yes indeed, I understand it! Many thanks!
$endgroup$
– bob
Jun 22 '17 at 17:48
add a comment |
$begingroup$
Using no contour integration we have:
We write $x=-|x|leq 0$, $int^{infty}_{0}frac{t^{a-1}}{t-x}dt=int^{infty}_{0}frac{t^{a-1}}{|x|+t}dt$.
Set first $t=w|x|$. Then the integral becomes
$$
|x|^{a-1}int^{infty}_{0}frac{w^{a-1}}{1+w}dw
$$
Set now $w=frac{y}{1-y}$. Hence the integral becomes
$$
|x|^{a-1}int^{1}_{0}y^{a-1}(1-y)^{-a}dy=|x|^{a-1}frac{Gamma(a)Gamma(1-a)}{Gamma(1)}=|x|^{a-1}Gamma(a)Gamma(1-a)
$$
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since the integrand is singular at $t=x$, this must be understood in the Cauchy principal value sense. Your contour should include little detours around $x=a$ as well as $0$.
The integrals over the arcs around $x$ are not $0$. For the top part, I get $-i pi x^{a-1}$; for the bottom part, $-i pi x^{a-1} e^{2 pi i (a-1)}$.
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
Oh yes indeed, I understand it! Many thanks!
$endgroup$
– bob
Jun 22 '17 at 17:48
add a comment |
$begingroup$
Since the integrand is singular at $t=x$, this must be understood in the Cauchy principal value sense. Your contour should include little detours around $x=a$ as well as $0$.
The integrals over the arcs around $x$ are not $0$. For the top part, I get $-i pi x^{a-1}$; for the bottom part, $-i pi x^{a-1} e^{2 pi i (a-1)}$.
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
Oh yes indeed, I understand it! Many thanks!
$endgroup$
– bob
Jun 22 '17 at 17:48
add a comment |
$begingroup$
Since the integrand is singular at $t=x$, this must be understood in the Cauchy principal value sense. Your contour should include little detours around $x=a$ as well as $0$.
The integrals over the arcs around $x$ are not $0$. For the top part, I get $-i pi x^{a-1}$; for the bottom part, $-i pi x^{a-1} e^{2 pi i (a-1)}$.
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
$endgroup$
Since the integrand is singular at $t=x$, this must be understood in the Cauchy principal value sense. Your contour should include little detours around $x=a$ as well as $0$.
The integrals over the arcs around $x$ are not $0$. For the top part, I get $-i pi x^{a-1}$; for the bottom part, $-i pi x^{a-1} e^{2 pi i (a-1)}$.
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
edited Jun 22 '17 at 17:28
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
answered Jun 22 '17 at 17:18
Robert IsraelRobert Israel
329k23217470
329k23217470
$begingroup$
Oh yes indeed, I understand it! Many thanks!
$endgroup$
– bob
Jun 22 '17 at 17:48
add a comment |
$begingroup$
Oh yes indeed, I understand it! Many thanks!
$endgroup$
– bob
Jun 22 '17 at 17:48
$begingroup$
Oh yes indeed, I understand it! Many thanks!
$endgroup$
– bob
Jun 22 '17 at 17:48
$begingroup$
Oh yes indeed, I understand it! Many thanks!
$endgroup$
– bob
Jun 22 '17 at 17:48
add a comment |
$begingroup$
Using no contour integration we have:
We write $x=-|x|leq 0$, $int^{infty}_{0}frac{t^{a-1}}{t-x}dt=int^{infty}_{0}frac{t^{a-1}}{|x|+t}dt$.
Set first $t=w|x|$. Then the integral becomes
$$
|x|^{a-1}int^{infty}_{0}frac{w^{a-1}}{1+w}dw
$$
Set now $w=frac{y}{1-y}$. Hence the integral becomes
$$
|x|^{a-1}int^{1}_{0}y^{a-1}(1-y)^{-a}dy=|x|^{a-1}frac{Gamma(a)Gamma(1-a)}{Gamma(1)}=|x|^{a-1}Gamma(a)Gamma(1-a)
$$
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
add a comment |
$begingroup$
Using no contour integration we have:
We write $x=-|x|leq 0$, $int^{infty}_{0}frac{t^{a-1}}{t-x}dt=int^{infty}_{0}frac{t^{a-1}}{|x|+t}dt$.
Set first $t=w|x|$. Then the integral becomes
$$
|x|^{a-1}int^{infty}_{0}frac{w^{a-1}}{1+w}dw
$$
Set now $w=frac{y}{1-y}$. Hence the integral becomes
$$
|x|^{a-1}int^{1}_{0}y^{a-1}(1-y)^{-a}dy=|x|^{a-1}frac{Gamma(a)Gamma(1-a)}{Gamma(1)}=|x|^{a-1}Gamma(a)Gamma(1-a)
$$
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
add a comment |
$begingroup$
Using no contour integration we have:
We write $x=-|x|leq 0$, $int^{infty}_{0}frac{t^{a-1}}{t-x}dt=int^{infty}_{0}frac{t^{a-1}}{|x|+t}dt$.
Set first $t=w|x|$. Then the integral becomes
$$
|x|^{a-1}int^{infty}_{0}frac{w^{a-1}}{1+w}dw
$$
Set now $w=frac{y}{1-y}$. Hence the integral becomes
$$
|x|^{a-1}int^{1}_{0}y^{a-1}(1-y)^{-a}dy=|x|^{a-1}frac{Gamma(a)Gamma(1-a)}{Gamma(1)}=|x|^{a-1}Gamma(a)Gamma(1-a)
$$
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
Using no contour integration we have:
We write $x=-|x|leq 0$, $int^{infty}_{0}frac{t^{a-1}}{t-x}dt=int^{infty}_{0}frac{t^{a-1}}{|x|+t}dt$.
Set first $t=w|x|$. Then the integral becomes
$$
|x|^{a-1}int^{infty}_{0}frac{w^{a-1}}{1+w}dw
$$
Set now $w=frac{y}{1-y}$. Hence the integral becomes
$$
|x|^{a-1}int^{1}_{0}y^{a-1}(1-y)^{-a}dy=|x|^{a-1}frac{Gamma(a)Gamma(1-a)}{Gamma(1)}=|x|^{a-1}Gamma(a)Gamma(1-a)
$$
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
answered Dec 31 '18 at 23:34
Nikos Bagis Nikos Bagis
2,482616
2,482616
add a comment |
add a comment |
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$begingroup$
This integral must be computed in the Cauchy sense, otehrwise it is undefined because of the pole at $t=x$.
$endgroup$
– Yves Daoust
Jun 22 '17 at 16:59
$begingroup$
WA gives the result containing a hypergeometric function
$endgroup$
– Dr. Sonnhard Graubner
Jun 22 '17 at 17:01
$begingroup$
@YvesDaoust But the pole $t=x$ is not in the contour. Or what am I doing wrong?
$endgroup$
– bob
Jun 22 '17 at 17:05
$begingroup$
The pole is absolutely on the real axis. This integral must be computed in the Cauchy sense, otherwise it is undefined.
$endgroup$
– Yves Daoust
Jun 22 '17 at 17:24
$begingroup$
@Dr.SonnhardGraubner What on earth are you thinking? The answer can be found in closed form.
$endgroup$
– Mark Viola
Jun 23 '17 at 3:03