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(1)

I want to compute $\int_{-\infty}^{+\infty}\frac{\cos x}{x^2+a^2}=\frac{\pi}{a}e^{-a}, a>0$

I started with partial integration, which gives me $$\int_{-\infty}^{+\infty}\frac{\cos x}{x^2+a^2}=\frac{\sin x}{x^2+a^2} |_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\frac{2x\sin x}{(x^2+a^2)^2}dx, $$

My idea was it might has something to do with $\int_0^\infty\frac{\sin x}{x}=\pi/2$ but I do not see any relation, may you can help me with that

(2) $\int_{-\infty}^{+\infty} \frac{x-\sin x}{x^3}dx=\pi/2$ I started by deforming the contour to avoid the origin and split the sin into exponentials: $\frac{x-\sin x}{x^3}=\frac{1}{x^2}+\frac{ie^{ix}}{2x^3}-\frac{ie^{-ix}}{2x^3}$, the residues of the three terms are 0,-1/4,-1/4, but this gives me the result $0$, something must be wrong here?

TI Jones
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1 Answers1

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For (a), I'm not sure what's preventing you from using the straightforward evaluation of

$$\int_{-\infty}^{\infty} dx \frac{e^{i x}}{x^2+a^2}$$

which may be attacked by considering

$$\oint_C dz \frac{e^{i z}}{z^2+a^2}$$

where $C$ is a semicircular contour in the upper half-plane or radius $R$. The integral over the circular arc vanishes as $\pi/R^3$ as $R \to\infty$, so that

$$\int_{-\infty}^{\infty} dx \frac{e^{i x}}{x^2+a^2} = i 2 \pi \frac{e^{i (i a)}}{2 i a} = \frac{\pi}{a} e^{-a}$$

The integral you seek is the real part of this, which is the same.

As for (b), you have to remember that, for the $e^{-i z}$ component, you close in the lower half plane, so the integral is $-i 2 \pi$ and thus the sum of the residues does not vanish.

Ron Gordon
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