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Calculate complex integral $int_{-infty}^infty e^{-ix^2}d x=?$

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2

$begingroup$

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 easy to handle either.

integration complex-analysis

share|cite|improve this question

edited 57 secs ago

Andrews

5891318

asked 3 hours ago

CollinCollin

1377

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – DavidG
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
  $endgroup$
  – DavidG
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming the jump into the complex domain is valid. I always do.
  $endgroup$
  – marty cohen
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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 easy to handle either.

integration complex-analysis

share|cite|improve this question

edited 57 secs ago

Andrews

5891318

asked 3 hours ago

CollinCollin

1377

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – DavidG
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
  $endgroup$
  – DavidG
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming the jump into the complex domain is valid. I always do.
  $endgroup$
  – marty cohen
  1 hour 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 easy to handle either.

integration complex-analysis

share|cite|improve this question

edited 57 secs ago

Andrews

5891318

asked 3 hours ago

CollinCollin

1377

$endgroup$

share|cite|improve this question

edited 57 secs ago

Andrews

5891318

asked 3 hours ago

CollinCollin

1377

share|cite|improve this question

edited 57 secs ago

Andrews

5891318

asked 3 hours ago

CollinCollin

1377

edited 57 secs ago

Andrews

5891318

edited 57 secs ago

Andrews

5891318

asked 3 hours ago

CollinCollin

1377

asked 3 hours ago

CollinCollin

1377

3 Answers
3

active

oldest

votes

5

$begingroup$

Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. 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 $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
  $endgroup$
  – Sangchul Lee
  38 mins ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

answered 1 hour ago

marty cohenmarty cohen

73.9k549128

$endgroup$

add a comment | 

2

$begingroup$

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?

share|cite|improve this answer

answered 1 hour ago

csch2csch2

2571311

$endgroup$

add a comment | 

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5

$begingroup$

Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. 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 $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
  $endgroup$
  – Sangchul Lee
  38 mins ago
  
  
  

  
  
  
  

add a comment | 

5

$begingroup$

Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. 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 $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
  $endgroup$
  – Sangchul Lee
  38 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

122k1157134

answered 2 hours ago

Claude LeiboviciClaude Leibovici

122k1157134

answered 2 hours ago

Claude LeiboviciClaude Leibovici

122k1157134

3

$begingroup$

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.

share|cite|improve this answer

answered 1 hour ago

marty cohenmarty cohen

73.9k549128

$endgroup$

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

answered 1 hour ago

marty cohenmarty cohen

73.9k549128

$endgroup$

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.

share|cite|improve this answer

answered 1 hour ago

marty cohenmarty cohen

73.9k549128

$endgroup$

share|cite|improve this answer

answered 1 hour ago

marty cohenmarty cohen

73.9k549128

answered 1 hour ago

marty cohenmarty cohen

73.9k549128

answered 1 hour ago

marty cohenmarty cohen

73.9k549128

2

$begingroup$

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?

share|cite|improve this answer

answered 1 hour ago

csch2csch2

2571311

$endgroup$

add a comment | 

2

$begingroup$

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?

share|cite|improve this answer

answered 1 hour ago

csch2csch2

2571311

$endgroup$

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?

share|cite|improve this answer

answered 1 hour ago

csch2csch2

2571311

$endgroup$

share|cite|improve this answer

answered 1 hour ago

csch2csch2

2571311

answered 1 hour ago

csch2csch2

2571311

answered 1 hour ago

csch2csch2

2571311