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Could you tell me how to solve this integral?

$$\int_0^{\infty} \frac{\cos x -1}{x^2}dx$$

I think I should focus on this integral $$\int_{\Gamma} \frac{e^{iz}-1}{z^2+ \varepsilon^2}$$

where $\Gamma$ is a curve = semicircle with radius $R \cup $ segment $[-R, R]$

The integral vanishes on the semicircle.

The poles of $\frac{e^{iz}-1}{z^2+ \varepsilon^2}$ are $\varepsilon i, - \varepsilon i$.

So $$\int_{\Gamma} \frac{e^{iz}-1}{z^2+ \varepsilon^2} = 2 \pi i \cdot res_{\varepsilon i} \left(\frac{e^{iz}-1}{z^2+ \varepsilon}\right) = 2 \pi i \frac{e^{- \varepsilon}-1}{2 \varepsilon i} = \frac{\pi (e^{- \varepsilon}-1)}{\varepsilon}$$

We have $\frac{0}{0}$ situation, so we can use de l'Hospital theorem:

$$\lim_{\varepsilon \rightarrow 0} \frac{\pi (e^{- \varepsilon}-1)}{\varepsilon} = \lim _{\epsilon \rightarrow 0}\pi \frac{- \varepsilon e^{- \varepsilon}}{1}$$

But the answer is $\frac{- \pi}{2}$

Could you tell me what I'm doing wrong?

Amith
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1 Answers1

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There is no extra $\varepsilon$ in the last limit. De l'Hopital rule just gives $-\pi$ as the value of the limit.

This limit is twice the integral since $$\int_{0}^{+\infty}\frac{\cos x-1}{x^2}\,dx = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{\cos x-1}{x^2}\,dx.$$

Also notice that this problem can also be solved through integration by parts: $$I=\int_{0}^{+\infty}\frac{\cos x-1}{x^2}\,dx=-\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=-\frac{\pi}{2}.$$

Jack D'Aurizio
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