Complex integral containing Gaussian

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I want to calculate the integral using complex integration:
$$ f(t) = int_-infty^infty dfrace^-(z+iat)^24z^2+1 dz = int_-infty^infty dfrace^-(z+iat)^2(2z-i)(2z+i) dz$$
where $a$ is real.



There are poles at $z=pm i/2$. If I draw a semicircle anticlockwise contour on the upper half of the complex plane, the contour encloses the $i/2$ pole. Then by the method of residues,



$$f(t) = 2pi imathrmRes_z=i/2 = 2pi i left( dfrac12i e^(1/2+at)^2right)$$



I double checked on Mathematica and this is not the correct answer.



I suspect I need to do something with the $e^-(z+iat)^2$ part, but I'm not sure what (I'm a novice at solving complex integrals). Can you please help?




complex-analysis residue-calculus complex-integration




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  • Hi Medulla Oblongata, what are the signs of $a,t$?
    – Szeto
    Jul 24 at 9:58










  • Hello again, Szeto. Both $a$ and $t$ are real, $t$ is positive and negative but $a$ can be restricted to being positive if necessary.
    – Medulla Oblongata
    Jul 24 at 10:01














up vote
0
down vote

favorite












I want to calculate the integral using complex integration:
$$ f(t) = int_-infty^infty dfrace^-(z+iat)^24z^2+1 dz = int_-infty^infty dfrace^-(z+iat)^2(2z-i)(2z+i) dz$$
where $a$ is real.



There are poles at $z=pm i/2$. If I draw a semicircle anticlockwise contour on the upper half of the complex plane, the contour encloses the $i/2$ pole. Then by the method of residues,



$$f(t) = 2pi imathrmRes_z=i/2 = 2pi i left( dfrac12i e^(1/2+at)^2right)$$



I double checked on Mathematica and this is not the correct answer.



I suspect I need to do something with the $e^-(z+iat)^2$ part, but I'm not sure what (I'm a novice at solving complex integrals). Can you please help?




complex-analysis residue-calculus complex-integration




share|cite|improve this question





















  • Hi Medulla Oblongata, what are the signs of $a,t$?
    – Szeto
    Jul 24 at 9:58










  • Hello again, Szeto. Both $a$ and $t$ are real, $t$ is positive and negative but $a$ can be restricted to being positive if necessary.
    – Medulla Oblongata
    Jul 24 at 10:01












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I want to calculate the integral using complex integration:
$$ f(t) = int_-infty^infty dfrace^-(z+iat)^24z^2+1 dz = int_-infty^infty dfrace^-(z+iat)^2(2z-i)(2z+i) dz$$
where $a$ is real.



There are poles at $z=pm i/2$. If I draw a semicircle anticlockwise contour on the upper half of the complex plane, the contour encloses the $i/2$ pole. Then by the method of residues,



$$f(t) = 2pi imathrmRes_z=i/2 = 2pi i left( dfrac12i e^(1/2+at)^2right)$$



I double checked on Mathematica and this is not the correct answer.



I suspect I need to do something with the $e^-(z+iat)^2$ part, but I'm not sure what (I'm a novice at solving complex integrals). Can you please help?




complex-analysis residue-calculus complex-integration




share|cite|improve this question













I want to calculate the integral using complex integration:
$$ f(t) = int_-infty^infty dfrace^-(z+iat)^24z^2+1 dz = int_-infty^infty dfrace^-(z+iat)^2(2z-i)(2z+i) dz$$
where $a$ is real.



There are poles at $z=pm i/2$. If I draw a semicircle anticlockwise contour on the upper half of the complex plane, the contour encloses the $i/2$ pole. Then by the method of residues,



$$f(t) = 2pi imathrmRes_z=i/2 = 2pi i left( dfrac12i e^(1/2+at)^2right)$$



I double checked on Mathematica and this is not the correct answer.



I suspect I need to do something with the $e^-(z+iat)^2$ part, but I'm not sure what (I'm a novice at solving complex integrals). Can you please help?





complex-analysis residue-calculus complex-integration





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share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 8:49
























asked Jul 24 at 8:39









Medulla Oblongata

271110




271110











  • Hi Medulla Oblongata, what are the signs of $a,t$?
    – Szeto
    Jul 24 at 9:58










  • Hello again, Szeto. Both $a$ and $t$ are real, $t$ is positive and negative but $a$ can be restricted to being positive if necessary.
    – Medulla Oblongata
    Jul 24 at 10:01
















  • Hi Medulla Oblongata, what are the signs of $a,t$?
    – Szeto
    Jul 24 at 9:58










  • Hello again, Szeto. Both $a$ and $t$ are real, $t$ is positive and negative but $a$ can be restricted to being positive if necessary.
    – Medulla Oblongata
    Jul 24 at 10:01















Hi Medulla Oblongata, what are the signs of $a,t$?
– Szeto
Jul 24 at 9:58




Hi Medulla Oblongata, what are the signs of $a,t$?
– Szeto
Jul 24 at 9:58












Hello again, Szeto. Both $a$ and $t$ are real, $t$ is positive and negative but $a$ can be restricted to being positive if necessary.
– Medulla Oblongata
Jul 24 at 10:01




Hello again, Szeto. Both $a$ and $t$ are real, $t$ is positive and negative but $a$ can be restricted to being positive if necessary.
– Medulla Oblongata
Jul 24 at 10:01










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How did you get $-1+i$? The residue at $z=i/2$ is $lim_zto i/2 (z-i/2) frac e^-(z+iat)^2 (2z-i)(2z+i)=frac 1 4i e^(1/2+at)^2$ and you have to multiply this by $2 pi i$






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  • thanks, I fixed the error
    – Medulla Oblongata
    Jul 24 at 8:49










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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













How did you get $-1+i$? The residue at $z=i/2$ is $lim_zto i/2 (z-i/2) frac e^-(z+iat)^2 (2z-i)(2z+i)=frac 1 4i e^(1/2+at)^2$ and you have to multiply this by $2 pi i$






share|cite|improve this answer





















  • thanks, I fixed the error
    – Medulla Oblongata
    Jul 24 at 8:49














up vote
0
down vote













How did you get $-1+i$? The residue at $z=i/2$ is $lim_zto i/2 (z-i/2) frac e^-(z+iat)^2 (2z-i)(2z+i)=frac 1 4i e^(1/2+at)^2$ and you have to multiply this by $2 pi i$






share|cite|improve this answer





















  • thanks, I fixed the error
    – Medulla Oblongata
    Jul 24 at 8:49












up vote
0
down vote










up vote
0
down vote









How did you get $-1+i$? The residue at $z=i/2$ is $lim_zto i/2 (z-i/2) frac e^-(z+iat)^2 (2z-i)(2z+i)=frac 1 4i e^(1/2+at)^2$ and you have to multiply this by $2 pi i$






share|cite|improve this answer













How did you get $-1+i$? The residue at $z=i/2$ is $lim_zto i/2 (z-i/2) frac e^-(z+iat)^2 (2z-i)(2z+i)=frac 1 4i e^(1/2+at)^2$ and you have to multiply this by $2 pi i$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 24 at 8:45









Kavi Rama Murthy

20.2k2829




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  • thanks, I fixed the error
    – Medulla Oblongata
    Jul 24 at 8:49
















  • thanks, I fixed the error
    – Medulla Oblongata
    Jul 24 at 8:49















thanks, I fixed the error
– Medulla Oblongata
Jul 24 at 8:49




thanks, I fixed the error
– Medulla Oblongata
Jul 24 at 8:49












 

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