I have $$\int_{-\infty}^{\infty}\frac{x^3sin(x)}{x^4+16}dx = \pi e^{-\sqrt{2}}cos(\sqrt{2})$$ and have been asked to show this using contour integration.

I have chosen the semicircular contour along the real axis from -R to R with the semicircle above the real axis. I have also made $$f(z)=\frac{z^3e^{iz}}{z^4+16}$$

I have found there to be poles at $$z=2e^{\frac{i\pi}{4}(1+2k)},\qquad k=0,1,2,3$$

And have tried showing that the integral across my contour is equal to $$2\pi i (res(f,2e^{\frac{\pi i}{4}})+res(f,2e^{\frac{3\pi i}{4}}))$$

Is this the correct choice of contour as I am having a lot of difficulty calculating the value of the residues? If it is the correct choice, how can we calculate the residues' values?

  • $\begingroup$ Seems OK. If you show your residue calculations we might also tell what (is something) goes wrong. $\endgroup$ – mickep Dec 5 '15 at 9:06
  • $\begingroup$ what exactly goes wrong with ur residues? everything else seems alright, maybe u have to be a little careful about how to justify the vanishing of the big arc. $\endgroup$ – tired Dec 5 '15 at 11:11

Suppose we seek to evaluate

$$\int_{-\infty}^\infty \frac{x^3\sin x}{x^4+16} dx$$

which is

$$\Im \int_{-\infty}^\infty \frac{x^3 \exp(ix)}{x^4+16} dx$$


$$f(z) = \frac{z^3 \exp(iz)}{z^4+16}$$

and integrating along a semicircular contour in the upper half plane consisting of a semicircle $\Gamma_1$ of radius $R$ and a segment $\Gamma_2$ on the real axis.

Following standard procedure we parameterize the integral along $\Gamma_1$ as $R\exp(i\theta)$ with $0\le\theta\le\pi$ to get

$$\left|\int_{\Gamma_1} \frac{z^3 \exp(iz)}{z^4+16} \; dz\right| \le \frac{R^3}{R^4-16} \int_0^{\pi} |\exp(iR\exp(i\theta))| \times |Ri \exp(i\theta)|\; d\theta \\ = \frac{R^4}{R^4-16} \int_0^{\pi} |\exp(iR\exp(i\theta))| \; d\theta .$$

Now using the symmetry of the sine and the bound for $0\le x\le \pi/2$ $$\sin x \ge \frac{2}{\pi} x$$

and $$|\exp(i R \exp(i\theta))| = |\exp(i R\cos\theta - R\sin\theta)| = \exp(-R\sin\theta).$$

we get for the remaining integral $$\int_0^{\pi} \exp(-R\sin\theta)\; d\theta \lt 2\int_0^{\pi/2} \exp(-R\theta 2/\pi) \; d\theta = -2\left[\frac{\pi}{2R} \exp(-R\theta 2/\pi)\right]_0^{\pi/2} \\ = \frac{\pi}{R} (1-\exp(-R)).$$

This yields for the integral along $\Gamma_1$ the bound $$\frac{\pi R^3}{R^4-16} (1-\exp(-R)) \rightarrow 0 \quad\text{as}\quad R\rightarrow \infty.$$

This vanishes as $R$ goes to infinity. It remains to sum the residues.

The poles are at $\rho_k$ with $0\le k\lt 4$ $$\rho_k = 2 \exp(i\pi /4 + \pi i k/2).$$ and we see that only $\rho_{0,1}$ are in the upper half plane.

We get $$\mathrm{Res}_{z=\rho_{0,1}} f(z) = \left. \frac{z^3 \exp(iz)}{4z^3}\right|_{z=\rho_{0,1}} = \left. \frac{1}{4} \exp(iz)\right|_{z=\rho_{0,1}}.$$

This yields for the complex integral the value $$2\pi i \times \frac{1}{4} (\exp(2i\exp(i\pi/4))+\exp(2i\exp(3i\pi/4)) \\ = 2\pi i \times \frac{1}{4} (\exp(\sqrt{2}i(1+i))+\exp(\sqrt{2}i(-1+i))) \\ = 2\pi i \times \frac{1}{4} (\exp(-\sqrt{2} + \sqrt{2}i)+\exp(-\sqrt{2} - \sqrt{2}i)) \\ = \pi i \times \frac{1}{2} \exp(-\sqrt{2}) (\exp(\sqrt{2}i)+\exp(- \sqrt{2}i)) \\ = \pi i \times \exp(-\sqrt{2}) \cos(\sqrt{2}).$$

Extracting the imaginary part we finally obtain $$\pi \exp(-\sqrt{2}) \cos(\sqrt{2}).$$

This computation was essentially the same as this MSE link.

| cite | improve this answer | |
  • $\begingroup$ Could you elaborate on the calculation of the residue? $\endgroup$ – John Keeper Jun 17 at 3:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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