# Solve $\int_{-\infty}^{\infty}\frac{x^3sin(x)}{x^4+16}dx$ using contour integration

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?

• Seems OK. If you show your residue calculations we might also tell what (is something) goes wrong. – mickep Dec 5 '15 at 9:06
• 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. – tired Dec 5 '15 at 11:11

## 1 Answer

Suppose we seek to evaluate

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

which is

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

using

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

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.

• Could you elaborate on the calculation of the residue? – John Keeper Jun 17 at 3:48