Evaluate the integral $$\int_0^{\infty}\frac{dx}{(x^2+25)(2+\cos x)}.$$
I know how to integrate
both the integrals
$$\int_0^{\infty}\frac{dx}{(x^2+25) }, \ \ \ \int_0^{\infty}\frac{dx}{2+\cos x}.$$
Is it any way to utilize the above integrals or please tell me how to proceed?
Evaluate the following importer intergal
calculusdefinite integralsimproper-integralsintegration
Best Answer
Let $\alpha = -2 + \sqrt{3}$, $\beta = -2 - \sqrt{3}$ and $u = e^{ix}$. For real $x$, we have:
$$\begin{align} \frac{1}{2 + \cos x} &= \frac{1}{2 + \frac{u + u^{-1}}{2}} = \frac{2u}{u^2+4u+1}\\ &= \frac{2u}{(u-\alpha)(u-\beta)} = \frac{2}{\alpha-\beta}\left(\frac{\alpha}{u-\alpha} - \frac{\beta}{u-\beta}\right)\\ &= \frac{1}{\sqrt{3}} \Re\left(\frac{\alpha}{u-\alpha} - \frac{\beta}{u-\beta}\right) = \frac{1}{\sqrt{3}} \Re\left(\frac{\alpha}{\bar{u}-\alpha} - \frac{\beta}{u-\beta}\right)\\ &= \frac{1}{\sqrt{3}} \Re\left(\frac{\beta^{-1}}{u^{-1}-\beta^{-1}} - \frac{\beta}{u-\beta}\right)\\ &= \frac{1}{\sqrt{3}}\Re\left(\frac{\beta + e^{ix}}{\beta - e^{ix}}\right) \end{align} $$ Using this, we can transform the integral at hand as
$$\begin{align} \mathcal{I} \stackrel{def}{=} &\; \int_0^\infty \frac{dx}{(x^2 + 5^2)(2+\cos x)} = \frac12 \int_{-\infty}^\infty \frac{dx}{(x^2 + 5^2)(2+\cos x)}\\ = &\;\frac{1}{2\sqrt{3}}\Re\int_{-\infty}^{\infty} \frac{1}{z^2+5^2}\frac{\beta + e^{iz}}{\beta - e^{iz}}dz \end{align} $$ Notice $|\beta| > 1$ and $|e^{iz}| \le 1$ for $\Im z \ge 0$. The factor $\frac{\beta + e^{iz}}{\beta - e^{iz}}$ is holomorphic and bounded over upper half plane. We can evaluate $\mathcal{I}$ by completing the contour in upper half plane and then take the residue at $5i$, the only pole of the integrand. The end result is
$$\begin{align} \mathcal{I} &= \frac{1}{2\sqrt{3}} \Re\left(2\pi i \mathop{\rm Res}\limits_{z=5i}\left[\frac{1}{z^2+5^2}\frac{\beta + e^{iz}}{\beta-e^{iz}}\right]\right) = \frac{1}{2\sqrt{3}}\Re\left(\frac{2\pi i}{10i}\frac{\beta + e^{-5}}{\beta - e^{-5}}\right)\\ &= \frac{\pi}{10\sqrt{3}}\frac{2+\sqrt{3} - e^{-5}}{2+\sqrt{3} + e^{-5}} \approx 0.180726180098288\ldots \end{align} $$