Techniques to solve this integral?

Question

$$\int_{-\infty}^{\infty}\frac{\cos x}{4x^2-\pi^2}dx$$ The question is from the Kyoto University exam, and I think it could be solved by the Residue Theorem, but I'm not familiar with it. I also tried Feynman Integral Method but it still doesn't work well.

Answer 1

You can decompose the integral as follows: $$\int_{-\infty}^\infty \frac{\cos(x)}{4x^2-\pi^2} \text{d}x = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\cos(x)}{2x-\pi} \text{d}x - \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\cos(x)}{2x+\pi} \text{d}x$$ Take the change of variable $x\to \tfrac{\pi}{2} - x$ in the first integral and $x\to x - \tfrac{\pi}{2}$ in the second: $$\int_{-\infty}^\infty \frac{\cos(x)}{4x^2-\pi^2} \text{d}x = -\frac{1}{4\pi}\int_{-\infty}^\infty \frac{\sin(x)}{x} \text{d}x - \frac{1}{4\pi}\int_{-\infty}^\infty \frac{\sin(x)}{x} \text{d}x = - \frac{1}{2\pi}\int_{-\infty}^\infty \frac{\sin(x)}{x} \text{d}x = -\frac{1}{2}$$ where I used the fact that $$\int_{-\infty}^\infty \frac{\sin(x)}{x} \text{d}x = \pi$$ as you can check several ways to solve this integral in here: Are there any ways to evaluate $\int^\infty_0\frac{\sin x}{x}dx$ without using double integral?

Answer 2

Rewriting the integral as $$ I=\int_{-\infty}^{\infty} \frac{\cos x}{4 x^2-\pi^2} d x=\Re\left[\int_{-\infty}^{\infty} \frac{e^{x i}}{4 x^2-\pi^2} d x\right], $$ then we can evaluate the integral using contour integration along anti-clockwise direction of the path $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi). $$

As $R\rightarrow \infty$, $$ \begin{aligned} I & = \Re \int_\gamma \frac{e^{z i}}{4 z^2-\pi^2} d z\\&=\Re\left[2 \pi i\left[\operatorname{Res}\left(\frac{e^{z i}}{4 z^2-\pi^2}, \pm \frac{\pi}{2}\right)\right]\right. \\ & =\Re\left[2 \pi i \left(\lim _{z\rightarrow \frac{\pi}{2}}\left(z-\frac{\pi}{2}\right) \frac{e^{z i}}{(2 z+\pi)(2 z-\pi)}+ \lim _{z\rightarrow \frac{-\pi}{2}}\left(z+\frac{\pi}{2}\right) \frac{e^{z i}}{(2 z+\pi)(2 z-\pi)}\right)\right] \\ & =\Re\left[\pi i\left(\frac{e^{\frac{\pi}{2}i} }{4 \pi}+\frac{e^{-\frac{\pi}{2} i}}{-4 \pi}\right)\right] \\ & =-\frac{1}{2} \end{aligned} $$

Answer 3

$$\frac{1}{4x^2-\pi^2}=\frac{1}{(2x-\pi)(2x+\pi)}=\frac 1{4\pi}\left( \frac 1{x-\frac \pi 2}-\frac 1{x+\frac \pi 2}\right)$$

Now, simple changes of variable $$\int\frac {\cos(x)}{x-\frac \pi 2}\,dx=\text{Si}\left(\frac{\pi }{2}-x\right)$$ $$\int\frac {\cos(x)}{x+\frac \pi 2}\,dx=\text{Si}\left(\frac{\pi }{2}+x\right)$$ $$\int \frac{\cos(x)}{4x^2-\pi^2}\,dx=\frac 1{4\pi}\left(\text{Si}\left(\frac{\pi }{2}-x\right)-\text{Si}\left(\frac{\pi }{2}+x\right)\right)$$ $$I=\int_{-t}^{+t} \frac{\cos(x)}{4x^2-\pi^2}\,dx=\frac 1{2\pi}\left(\text{Si}\left(\frac{\pi }{2}-t\right)-\text{Si}\left(\frac{\pi}{2}+t\right)\right)$$

Expanding for large values of $t$

$$I=-\frac{1}{2}+\frac{\sin (t)}{2 t^2}-\frac{\cos (t)}{t^3}+\cdots$$ which is a very good approximation as soon as $t \geq 2\pi$.