Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I evaluate $$\int_{-\infty}^\infty {\exp(ixk)\over -x^2+2ixa+a^2+b^2} dx,$$ where $k\in \mathbb R, a>0$? Would Fourier transforms simplify anything? I know very little about complex analysis, so I am guessing there is a rather simple way to evaluate this? Thanks.

share|cite|improve this question
up vote 5 down vote accepted

You can evaluate this using the residue theorem. The integrand has simple poles at $x_\pm=\mathrm ia\pm b$ (which you can find by setting the denominator zero and solving the quadratic equation). You can find the residues at the poles by multiplying the integrand by $x-x_\pm$ and then substituting $x_\pm$, which yields

$$\left.-\frac{\exp(\mathrm ixk)}{x-(\mathrm ia\mp b)}\right|_{x=\mathrm ia\pm b}=\mp\exp(-ka)\frac{\exp(\pm\mathrm ikb)}{2b}\;.$$

If $k\gt 0$, You can complete the integral by a half circle at infinity in the upper half-plane (which contains the poles), since the integrand decays quadratically with $x\to\pm\infty$ and exponentially with $x\to\mathrm i\infty$, so the contribution from this half circle vanishes. Thus by the residue theorem the given integral is $2\pi\mathrm i$ times the sum of the residues, that is,

$$2\pi\mathrm i\left(-\exp(-ka)\frac{\exp(\mathrm ikb)}{2b}+\exp(-ka)\frac{\exp(-\mathrm ikb)}{2b}\right)=2\pi\exp(-ka)\frac{\sin(kb)}b\;.$$

If $k\le0$, you can complete the integral by a half circle at infinity in the lower half-plane, since the integrand decays quadratically (or exponentially for $k\lt0$) and the perimeter of the half circle increases linearly. There are no poles in the lower half-plane, so in this case the integral is zero.

share|cite|improve this answer
Thanks, joriki! – rook Nov 16 '11 at 10:53
Basically the same remark as for Ali's answer. If $k<0$ the contour should be closed in the lower half-plane and the integral will be zero. – Heike Nov 16 '11 at 11:58
@Heike: You're right of course; I confused $a\gt0$ with $k\gt0$; I'll edit the answer accordingly. – joriki Nov 16 '11 at 12:08
Remains the case $b=0$. – Did Nov 16 '11 at 12:18

Assume $b \neq 0$ and $k\neq 0$.

Write $\dfrac{\exp(ixk)}{-x^2+2iax+a^2+b^2}= \dfrac{\exp(ixk)}{-(x-ia)^2+b^2}=\dfrac{\exp(i(x-ia)k)}{-(x-ia)^2+b^2} \exp(-ak)$ hence the integral becomes $I=\int_{-\infty}^\infty \dfrac{\exp(i(x-ia)k)}{-(x-ia)^2+b^2} \exp(-ak)dx=\int_{-\infty-ia}^{\infty -ia} \dfrac{\exp(izk)}{-z^2+b^2} \exp(-ak)dz$ on the contour the straight line parallel the $x$-axis and intercepting the $y$-axis (imaginary) at $-ia$. We need to close the contour by a great semicircle in the upper half plane if $k>0$ and in this case there are two poles at $z=b$ and $z=-b$ enclosed in the contour. Now we will use the residue theorem. The poles of the fraction $\dfrac{\exp(izk)}{-z^2+b^2}$ are $-b,+b$ and then the integral will be $I=(2\pi i)\exp(-ak)\lbrace Res(z=b)\exp(ibk)+Res(z=-b)\exp(-ibk)\rbrace$. Now observe that $Res(z=b)=1/2b$ and $Res(z=-b)=-1/2b$ hence $I=2 \pi i \exp(-ak) \frac{\sin bk}{b}$. If $k<0$ then we close the contour by a semicircle in the lower halfplane and in this case there are no poles enclosed and so the integral becomes zero.

Now assume $b=0$ and $K\neq 0$: in this case the residue (at $z=0$) becomes $-ik$ and the integral becomes $2\pi k$ (if $k>0$) and $0$ if $k<0$.

Finally let $k=0$: then in this case the result will be easy and I leave it for you as an exersice.

share|cite|improve this answer
Thank you, Ali! – rook Nov 16 '11 at 10:53
you have to be careful to close the contour in the right half-plane. Since $\exp(izk)$ blows up for $Re(izk)\rightarrow\infty$ and goes to zero for $Re(izk)\rightarrow-\infty$, you should close the contour in the upper half-plane for $k>0$ and in the lower half-plane for $k<0$. Depending on the sign of $k$ you'll have either 0 or 2 poles inside the closed contour. – Heike Nov 16 '11 at 11:50
Remains the case $b=0$. – Did Nov 16 '11 at 12:18
Thanks for your remarks. I will make the necessary changes later tonight. – user17090 Nov 16 '11 at 13:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.