How to calculate this simple integral?
$begingroup$
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.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
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 $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( 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 trying to vanish.
integration quantum-mechanics greens-function
$endgroup$
add a comment |
$begingroup$
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.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
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 $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( 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 trying to vanish.
integration quantum-mechanics greens-function
$endgroup$
$begingroup$
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.
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– DavidG
2 hours ago
1
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Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
2 hours ago
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
34 mins ago
add a comment |
$begingroup$
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.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
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 $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( 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 trying to vanish.
integration quantum-mechanics greens-function
$endgroup$
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.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
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 $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( 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 trying to vanish.
integration quantum-mechanics greens-function
integration quantum-mechanics greens-function
edited 18 mins ago
Collin
asked 2 hours ago
CollinCollin
1377
1377
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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
2 hours ago
1
$begingroup$
Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
2 hours ago
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
34 mins ago
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
2 hours ago
1
$begingroup$
Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
2 hours ago
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
34 mins ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
1
1
$begingroup$
Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
2 hours ago
$begingroup$
Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
2 hours ago
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
34 mins ago
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
34 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$
$endgroup$
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ 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
12 mins ago
add a comment |
$begingroup$
Trying to avoid complex funniness.
$begin{array}\
int_{-infty}^infty e^{-ix^2}dx
&=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\
end{array}
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfrac{sqrt{pi}}{8}
$
as $x to infty$.
Therefore the result is
$(1-i)sqrt{frac{pi}{2}}
$
as Claude Leibovici
got.
$endgroup$
add a comment |
$begingroup$
Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
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?
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$
$endgroup$
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ 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
12 mins ago
add a comment |
$begingroup$
Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$
$endgroup$
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ 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
12 mins ago
add a comment |
$begingroup$
Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$
$endgroup$
Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$
answered 1 hour ago
Claude LeiboviciClaude Leibovici
122k1157134
122k1157134
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ 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
12 mins ago
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 $mathbb{C}$ 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
12 mins ago
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ 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
12 mins ago
$begingroup$
Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ 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
12 mins ago
add a comment |
$begingroup$
Trying to avoid complex funniness.
$begin{array}\
int_{-infty}^infty e^{-ix^2}dx
&=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\
end{array}
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfrac{sqrt{pi}}{8}
$
as $x to infty$.
Therefore the result is
$(1-i)sqrt{frac{pi}{2}}
$
as Claude Leibovici
got.
$endgroup$
add a comment |
$begingroup$
Trying to avoid complex funniness.
$begin{array}\
int_{-infty}^infty e^{-ix^2}dx
&=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\
end{array}
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfrac{sqrt{pi}}{8}
$
as $x to infty$.
Therefore the result is
$(1-i)sqrt{frac{pi}{2}}
$
as Claude Leibovici
got.
$endgroup$
add a comment |
$begingroup$
Trying to avoid complex funniness.
$begin{array}\
int_{-infty}^infty e^{-ix^2}dx
&=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\
end{array}
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfrac{sqrt{pi}}{8}
$
as $x to infty$.
Therefore the result is
$(1-i)sqrt{frac{pi}{2}}
$
as Claude Leibovici
got.
$endgroup$
Trying to avoid complex funniness.
$begin{array}\
int_{-infty}^infty e^{-ix^2}dx
&=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\
end{array}
$
and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfrac{sqrt{pi}}{8}
$
as $x to infty$.
Therefore the result is
$(1-i)sqrt{frac{pi}{2}}
$
as Claude Leibovici
got.
answered 35 mins ago
marty cohenmarty cohen
73.9k549128
73.9k549128
add a comment |
add a comment |
$begingroup$
Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
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?
$endgroup$
add a comment |
$begingroup$
Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
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?
$endgroup$
add a comment |
$begingroup$
Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
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?
$endgroup$
Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
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 38 mins ago
csch2csch2
2571311
2571311
add a comment |
add a comment |
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$begingroup$
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
2 hours ago
1
$begingroup$
Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
2 hours ago
$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
34 mins ago