Calculate complex integral $int_-infty^infty e^-ix^2d x=?$ [duplicate]
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$begingroup$
This question already has an answer here:
Integration (Fourier transform)
2 answers
How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_-infty^infty e^-ix^2d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.
Thank you very much if you can help me out! And I would be grateful if you can give more than one approach
P.S.: The equation $int _-infty^inftye^-kt^2d sqrtkt=sqrtpi$ surely comes to my mind, but I don't know why it holds for $kinmathbbC$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^i pi/4x$ ( So I think it's the problem with my complex integral knowledge.)
I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.
integration complex-analysis
edited Feb 25 at 7:02
Andrews
1,2691421
asked Feb 25 at 3:31
CollinCollin
1447
$endgroup$
marked as duplicate by Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Lee David Chung Lin, Song, Leucippus Feb 27 at 1:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Integration (Fourier transform)
2 answers
How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_-infty^infty e^-ix^2d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.
Thank you very much if you can help me out! And I would be grateful if you can give more than one approach
P.S.: The equation $int _-infty^inftye^-kt^2d sqrtkt=sqrtpi$ surely comes to my mind, but I don't know why it holds for $kinmathbbC$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^i pi/4x$ ( So I think it's the problem with my complex integral knowledge.)
I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.
integration complex-analysis
edited Feb 25 at 7:02
Andrews
1,2691421
asked Feb 25 at 3:31
CollinCollin
1447
$endgroup$
marked as duplicate by Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Lee David Chung Lin, Song, Leucippus Feb 27 at 1:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
$endgroup$
– DavidG
Feb 25 at 3:46
1
$begingroup$
Hint: This is all you need to solve: $$ int_-infty^infty e^-x^2:dx = sqrtpi$$
$endgroup$
– DavidG
Feb 25 at 3:48
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
Feb 25 at 5:39
add a comment |
$begingroup$
This question already has an answer here:
Integration (Fourier transform)
2 answers
How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_-infty^infty e^-ix^2d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.
Thank you very much if you can help me out! And I would be grateful if you can give more than one approach
P.S.: The equation $int _-infty^inftye^-kt^2d sqrtkt=sqrtpi$ surely comes to my mind, but I don't know why it holds for $kinmathbbC$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^i pi/4x$ ( So I think it's the problem with my complex integral knowledge.)
I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.
integration complex-analysis
edited Feb 25 at 7:02
Andrews
1,2691421
asked Feb 25 at 3:31
CollinCollin
1447
$endgroup$
This question already has an answer here:
Integration (Fourier transform)
2 answers
How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_-infty^infty e^-ix^2d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.
Thank you very much if you can help me out! And I would be grateful if you can give more than one approach
P.S.: The equation $int _-infty^inftye^-kt^2d sqrtkt=sqrtpi$ surely comes to my mind, but I don't know why it holds for $kinmathbbC$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^i pi/4x$ ( So I think it's the problem with my complex integral knowledge.)
I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.
This question already has an answer here:
Integration (Fourier transform)
2 answers
integration complex-analysis
integration complex-analysis
edited Feb 25 at 7:02
Andrews
1,2691421
asked Feb 25 at 3:31
CollinCollin
1447
edited Feb 25 at 7:02
Andrews
1,2691421
asked Feb 25 at 3:31
CollinCollin
1447
edited Feb 25 at 7:02
Andrews
1,2691421
edited Feb 25 at 7:02
Andrews
1,2691421
edited Feb 25 at 7:02
Andrews
1,2691421
1,2691421
asked Feb 25 at 3:31
CollinCollin
1447
asked Feb 25 at 3:31
CollinCollin
1447
asked Feb 25 at 3:31
CollinCollin
1447
1447
marked as duplicate by Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Lee David Chung Lin, Song, Leucippus Feb 27 at 1:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, GNUSupporter 8964民主女神 地下教會, Lee David Chung Lin, Song, Leucippus Feb 27 at 1:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
$endgroup$
– DavidG
Feb 25 at 3:46
1
$begingroup$
Hint: This is all you need to solve: $$ int_-infty^infty e^-x^2:dx = sqrtpi$$
$endgroup$
– DavidG
Feb 25 at 3:48
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
Feb 25 at 5:39
add a comment |
$begingroup$
Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
$endgroup$
– DavidG
Feb 25 at 3:46
1
$begingroup$
Hint: This is all you need to solve: $$ int_-infty^infty e^-x^2:dx = sqrtpi$$
$endgroup$
– DavidG
Feb 25 at 3:48
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
Feb 25 at 5:39
$begingroup$
Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
$endgroup$
– DavidG
Feb 25 at 3:46
$begingroup$
Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
$endgroup$
– DavidG
Feb 25 at 3:46
1
1
$begingroup$
Hint: This is all you need to solve: $$ int_-infty^infty e^-x^2:dx = sqrtpi$$
$endgroup$
– DavidG
Feb 25 at 3:48
$begingroup$
Hint: This is all you need to solve: $$ int_-infty^infty e^-x^2:dx = sqrtpi$$
$endgroup$
– DavidG
Feb 25 at 3:48
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
Feb 25 at 5:39
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
Feb 25 at 5:39
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Trying to avoid complex funniness.
$beginarray\
int_-infty^infty e^-ix^2dx
&=int_-infty^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty cos(x^2)dx-2iint_0^inftysin(x^2))dx\
endarray
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfracsqrtpi8
$
as $x to infty$.
Therefore the result is
$(1-i)sqrtfracpi2
$
as Claude Leibovici
got.
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
How did you avoid 'complexness' in this solution? :-)
$endgroup$
– DavidG
Feb 25 at 9:41
$begingroup$
By not letting $k$ be complex in. $int e^-kx^2 dx$ when the result had only been proven for real $k$.
$endgroup$
– marty cohen
Feb 25 at 14:20
$begingroup$
Good way to avoid any further discussion about the complex ! Thanks.
$endgroup$
– Claude Leibovici
Feb 26 at 3:21
add a comment |
$begingroup$
Hint
$$int e^-k x^2,dx=fracsqrtpi 2 sqrtk,texterfleft(sqrtk xright)$$
$$f(k)=int_-infty^infty e^-k x^2,dx=fracsqrtpi sqrtk$$
$$f(i)=fracsqrtpi sqrti=(1-i) sqrtfracpi 2$$
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbbC$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
$endgroup$
– Collin
Feb 25 at 6:01
$begingroup$
@Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H=k:operatornameRe(k) > 0$ and a continuous function on $overlineHsetminus0 = k : operatornameRe(k) geq 0, k neq 0 $. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrtpi/k$, and so, $f = g$ on $H$ by identity theorem and on all of $overlineHsetminus0$ by continuity.
$endgroup$
– Sangchul Lee
Feb 25 at 6:24
$begingroup$
@SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter.
$endgroup$
– Collin
Feb 25 at 19:57
$begingroup$
See my answer that purposely avoids this problem.
$endgroup$
– marty cohen
Feb 26 at 0:03
add a comment |
$begingroup$
Hint:$$int_-infty^infty e^-kx^2dx=int_-infty^infty e^-left(xsqrt kright)^2dx$$
Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?
answered Feb 25 at 5:35
csch2csch2
6031314
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Trying to avoid complex funniness.
$beginarray\
int_-infty^infty e^-ix^2dx
&=int_-infty^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty cos(x^2)dx-2iint_0^inftysin(x^2))dx\
endarray
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfracsqrtpi8
$
as $x to infty$.
Therefore the result is
$(1-i)sqrtfracpi2
$
as Claude Leibovici
got.
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
How did you avoid 'complexness' in this solution? :-)
$endgroup$
– DavidG
Feb 25 at 9:41
$begingroup$
By not letting $k$ be complex in. $int e^-kx^2 dx$ when the result had only been proven for real $k$.
$endgroup$
– marty cohen
Feb 25 at 14:20
$begingroup$
Good way to avoid any further discussion about the complex ! Thanks.
$endgroup$
– Claude Leibovici
Feb 26 at 3:21
add a comment |
$begingroup$
Trying to avoid complex funniness.
$beginarray\
int_-infty^infty e^-ix^2dx
&=int_-infty^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty cos(x^2)dx-2iint_0^inftysin(x^2))dx\
endarray
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfracsqrtpi8
$
as $x to infty$.
Therefore the result is
$(1-i)sqrtfracpi2
$
as Claude Leibovici
got.
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
How did you avoid 'complexness' in this solution? :-)
$endgroup$
– DavidG
Feb 25 at 9:41
$begingroup$
By not letting $k$ be complex in. $int e^-kx^2 dx$ when the result had only been proven for real $k$.
$endgroup$
– marty cohen
Feb 25 at 14:20
$begingroup$
Good way to avoid any further discussion about the complex ! Thanks.
$endgroup$
– Claude Leibovici
Feb 26 at 3:21
add a comment |
$begingroup$
Trying to avoid complex funniness.
$beginarray\
int_-infty^infty e^-ix^2dx
&=int_-infty^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty cos(x^2)dx-2iint_0^inftysin(x^2))dx\
endarray
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfracsqrtpi8
$
as $x to infty$.
Therefore the result is
$(1-i)sqrtfracpi2
$
as Claude Leibovici
got.
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
$endgroup$
Trying to avoid complex funniness.
$beginarray\
int_-infty^infty e^-ix^2dx
&=int_-infty^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty (cos(x^2)-isin(x^2))dx\
&=2int_0^infty cos(x^2)dx-2iint_0^inftysin(x^2))dx\
endarray
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfracsqrtpi8
$
as $x to infty$.
Therefore the result is
$(1-i)sqrtfracpi2
$
as Claude Leibovici
got.
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
answered Feb 25 at 5:38
marty cohenmarty cohen
74.5k549129
74.5k549129
$begingroup$
How did you avoid 'complexness' in this solution? :-)
$endgroup$
– DavidG
Feb 25 at 9:41
$begingroup$
By not letting $k$ be complex in. $int e^-kx^2 dx$ when the result had only been proven for real $k$.
$endgroup$
– marty cohen
Feb 25 at 14:20
$begingroup$
Good way to avoid any further discussion about the complex ! Thanks.
$endgroup$
– Claude Leibovici
Feb 26 at 3:21
add a comment |
$begingroup$
How did you avoid 'complexness' in this solution? :-)
$endgroup$
– DavidG
Feb 25 at 9:41
$begingroup$
By not letting $k$ be complex in. $int e^-kx^2 dx$ when the result had only been proven for real $k$.
$endgroup$
– marty cohen
Feb 25 at 14:20
$begingroup$
Good way to avoid any further discussion about the complex ! Thanks.
$endgroup$
– Claude Leibovici
Feb 26 at 3:21
$begingroup$
How did you avoid 'complexness' in this solution? :-)
$endgroup$
– DavidG
Feb 25 at 9:41
$begingroup$
How did you avoid 'complexness' in this solution? :-)
$endgroup$
– DavidG
Feb 25 at 9:41
$begingroup$
By not letting $k$ be complex in. $int e^-kx^2 dx$ when the result had only been proven for real $k$.
$endgroup$
– marty cohen
Feb 25 at 14:20
$begingroup$
By not letting $k$ be complex in. $int e^-kx^2 dx$ when the result had only been proven for real $k$.
$endgroup$
– marty cohen
Feb 25 at 14:20
$begingroup$
Good way to avoid any further discussion about the complex ! Thanks.
$endgroup$
– Claude Leibovici
Feb 26 at 3:21
$begingroup$
Good way to avoid any further discussion about the complex ! Thanks.
$endgroup$
– Claude Leibovici
Feb 26 at 3:21
add a comment |
$begingroup$
Hint
$$int e^-k x^2,dx=fracsqrtpi 2 sqrtk,texterfleft(sqrtk xright)$$
$$f(k)=int_-infty^infty e^-k x^2,dx=fracsqrtpi sqrtk$$
$$f(i)=fracsqrtpi sqrti=(1-i) sqrtfracpi 2$$
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbbC$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
$endgroup$
– Collin
Feb 25 at 6:01
$begingroup$
@Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H=k:operatornameRe(k) > 0$ and a continuous function on $overlineHsetminus0 = k : operatornameRe(k) geq 0, k neq 0 $. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrtpi/k$, and so, $f = g$ on $H$ by identity theorem and on all of $overlineHsetminus0$ by continuity.
$endgroup$
– Sangchul Lee
Feb 25 at 6:24
$begingroup$
@SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter.
$endgroup$
– Collin
Feb 25 at 19:57
$begingroup$
See my answer that purposely avoids this problem.
$endgroup$
– marty cohen
Feb 26 at 0:03
add a comment |
$begingroup$
Hint
$$int e^-k x^2,dx=fracsqrtpi 2 sqrtk,texterfleft(sqrtk xright)$$
$$f(k)=int_-infty^infty e^-k x^2,dx=fracsqrtpi sqrtk$$
$$f(i)=fracsqrtpi sqrti=(1-i) sqrtfracpi 2$$
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbbC$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
$endgroup$
– Collin
Feb 25 at 6:01
$begingroup$
@Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H=k:operatornameRe(k) > 0$ and a continuous function on $overlineHsetminus0 = k : operatornameRe(k) geq 0, k neq 0 $. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrtpi/k$, and so, $f = g$ on $H$ by identity theorem and on all of $overlineHsetminus0$ by continuity.
$endgroup$
– Sangchul Lee
Feb 25 at 6:24
$begingroup$
@SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter.
$endgroup$
– Collin
Feb 25 at 19:57
$begingroup$
See my answer that purposely avoids this problem.
$endgroup$
– marty cohen
Feb 26 at 0:03
add a comment |
$begingroup$
Hint
$$int e^-k x^2,dx=fracsqrtpi 2 sqrtk,texterfleft(sqrtk xright)$$
$$f(k)=int_-infty^infty e^-k x^2,dx=fracsqrtpi sqrtk$$
$$f(i)=fracsqrtpi sqrti=(1-i) sqrtfracpi 2$$
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Hint
$$int e^-k x^2,dx=fracsqrtpi 2 sqrtk,texterfleft(sqrtk xright)$$
$$f(k)=int_-infty^infty e^-k x^2,dx=fracsqrtpi sqrtk$$
$$f(i)=fracsqrtpi sqrti=(1-i) sqrtfracpi 2$$
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
answered Feb 25 at 4:52
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbbC$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
$endgroup$
– Collin
Feb 25 at 6:01
$begingroup$
@Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H=k:operatornameRe(k) > 0$ and a continuous function on $overlineHsetminus0 = k : operatornameRe(k) geq 0, k neq 0 $. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrtpi/k$, and so, $f = g$ on $H$ by identity theorem and on all of $overlineHsetminus0$ by continuity.
$endgroup$
– Sangchul Lee
Feb 25 at 6:24
$begingroup$
@SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter.
$endgroup$
– Collin
Feb 25 at 19:57
$begingroup$
See my answer that purposely avoids this problem.
$endgroup$
– marty cohen
Feb 26 at 0:03
add a comment |
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbbC$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
$endgroup$
– Collin
Feb 25 at 6:01
$begingroup$
@Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H=k:operatornameRe(k) > 0$ and a continuous function on $overlineHsetminus0 = k : operatornameRe(k) geq 0, k neq 0 $. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrtpi/k$, and so, $f = g$ on $H$ by identity theorem and on all of $overlineHsetminus0$ by continuity.
$endgroup$
– Sangchul Lee
Feb 25 at 6:24
$begingroup$
@SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter.
$endgroup$
– Collin
Feb 25 at 19:57
$begingroup$
See my answer that purposely avoids this problem.
$endgroup$
– marty cohen
Feb 26 at 0:03
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Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbbC$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
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– Collin
Feb 25 at 6:01
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Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbbC$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
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– Collin
Feb 25 at 6:01
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@Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H=k:operatornameRe(k) > 0$ and a continuous function on $overlineHsetminus0 = k : operatornameRe(k) geq 0, k neq 0 $. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrtpi/k$, and so, $f = g$ on $H$ by identity theorem and on all of $overlineHsetminus0$ by continuity.
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– Sangchul Lee
Feb 25 at 6:24
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@Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H=k:operatornameRe(k) > 0$ and a continuous function on $overlineHsetminus0 = k : operatornameRe(k) geq 0, k neq 0 $. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrtpi/k$, and so, $f = g$ on $H$ by identity theorem and on all of $overlineHsetminus0$ by continuity.
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– Sangchul Lee
Feb 25 at 6:24
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@SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter.
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– Collin
Feb 25 at 19:57
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@SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter.
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– Collin
Feb 25 at 19:57
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See my answer that purposely avoids this problem.
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– marty cohen
Feb 26 at 0:03
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See my answer that purposely avoids this problem.
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– marty cohen
Feb 26 at 0:03
add a comment |
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Hint:$$int_-infty^infty e^-kx^2dx=int_-infty^infty e^-left(xsqrt kright)^2dx$$
Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?
answered Feb 25 at 5:35
csch2csch2
6031314
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add a comment |
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Hint:$$int_-infty^infty e^-kx^2dx=int_-infty^infty e^-left(xsqrt kright)^2dx$$
Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?
answered Feb 25 at 5:35
csch2csch2
6031314
$endgroup$
add a comment |
$begingroup$
Hint:$$int_-infty^infty e^-kx^2dx=int_-infty^infty e^-left(xsqrt kright)^2dx$$
Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?
answered Feb 25 at 5:35
csch2csch2
6031314
$endgroup$
Hint:$$int_-infty^infty e^-kx^2dx=int_-infty^infty e^-left(xsqrt kright)^2dx$$
Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?
answered Feb 25 at 5:35
csch2csch2
6031314
answered Feb 25 at 5:35
csch2csch2
6031314
answered Feb 25 at 5:35
csch2csch2
6031314
answered Feb 25 at 5:35
csch2csch2
6031314
6031314
add a comment |
add a comment |
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Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
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– DavidG
Feb 25 at 3:46
1
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Hint: This is all you need to solve: $$ int_-infty^infty e^-x^2:dx = sqrtpi$$
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– DavidG
Feb 25 at 3:48
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Assuming the jump into the complex domain is valid. I always do.
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– marty cohen
Feb 25 at 5:39