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How can I obtain the result of the following integration? $$\int_{-\infty}^{+\infty} \log \left(1+\frac{a^2}{x^2}\right)dx$$

Thank you!

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3 Answers 3

Integrating by parts

$$\int \log\left(1+\frac{a^2}{x^2}\right)dx= x\log\left(1+\frac{a^2}{x^2}\right)+2a^2\int\frac{1}{x^2+a^2}dx\\=x\log\left(1+\frac{a^2}{x^2}\right)+2a\tan^{-1}\left(\frac{x}{a}\right)+C$$


$$\lim_{x \rightarrow \infty}2a\tan^{-1}\left(\frac{x}{a}\right)=\pi a$$

we get

$$PV\int_{-\infty}^{\infty} \log\left(1+\frac{a^2}{x^2}\right)dx=2\pi a$$

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Use integration by parts.

$x log(1+\frac{a^2}{x^2})+ \int \frac {1}{1+\frac{a^2}{x^2}}.\frac{2a^2}{x^3}.x dx$

Answer: $x \log(1+\frac{a^2}{x^2})+ 2a \tan^{-1} \frac{x}{a}$ Put in the limits now!That will give you the value $2 \pi a$

Hope this helps.

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$$\int_{-\infty}^{\infty}\log \left(1+\frac{a^2}{x^2}\right)=x\log \left(1+\frac{a^2}{x^2}\right)|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\frac{-2a^2}{x^2+a^2}=2a\int_{-\infty}^{\infty}\tan^{-1} \frac{x}{a}=2a\pi$$

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