How do I evaluate this : $\int_{0}^{\infty} \ln \left( 1 + \frac{a^{2}}{x^{2}}\right)\ dx $ for $a > 0$?

Question

How do I evaluate this integral if I suppose that $a > 0$

$$\int_{0}^{\infty} \ln \left( 1 + \frac{a^{2}}{x^{2}}\right)\ \mathrm{d}x .$$

For $a=2$ I got $2\pi$ I think the result will be $a\pi$.

Answer 1

According to an answer by rae306 on Art of Problem Solving:

Using integration by parts: $$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx\,\,\begin{bmatrix}u=\ln\left(1+\frac{a^2}{x^2}\right)& du=\frac{\frac{-2a^2}{x}}{x^2+a^2}\,dx \\ dv=dx& v=x\end{bmatrix}\\=\underbrace{\left.x\cdot \ln\left(1+\frac{a^2}{x^2}\right)\right|_0^{\infty}}_{L}+\underbrace{\int_0^{\infty}\frac{2a^2}{x^2+a^2}\,dx}_{I}$$ $$I=\int_0^{\infty}\frac{2a^2}{x^2+a^2}\,dx=\left.2a^2\cdot\frac{1}{a}\arctan\left(\frac{x}{a}\right)\right\vert_0^\infty=2a\cdot\frac{\pi}{2}=\pi a$$ We now have to show that $L=0$: $$\lim_{x\to\infty}x\cdot\ln\left(1+\frac{a^2}{x^2}\right)=\lim_{x\to\infty}x\left(\frac{a^2}{x^2}-\frac{\left(\frac{a^2}{x^2}\right)^2}{2}+\frac{\left(\frac{a^2}{x^2}\right)^3}{3}-\frac{\left(\frac{a^2}{x^2}\right)^4}{4}+\ldots\right)\\=\lim_{x\to\infty}\frac{a^2}{x}-\frac{a^4}{2x^3}+\frac{a^6}{3x^5}-\frac{a^8}{4x^7}+\ldots=0$$ by using the series expansion. $$\lim_{x\to0}x\cdot\ln\left(1+\frac{a^2}{x^2}\right)=\lim_{x\to 0} \frac{\ln\left(1+\frac{a^2}{x^2}\right)}{\frac{1}{x}}=\lim_{x\to 0}\frac{\frac{\frac{2a^2}{x}}{x^2+a^2}}{\frac{1}{x^2}}=\lim_{x\to0}\frac{2a^2 x}{x^2+a^2}=0$$ using L'Hôpital's Rule. Therefore $$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx= a\pi$$ $\square$

Answer 2

$$\int_{0}^{+\infty}\log\left(1+\frac{a^2}{x^2}\right)\,dx = a\int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx $$ then by setting $x=\tan\theta$ we have:

$$ \int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx = -2\int_{0}^{\pi/2}\frac{\log\sin\theta}{\cos^2\theta}\,d\theta=2\int_{0}^{\pi/2}\cot(\theta)\tan(\theta)\,d\theta=\color{red}{\pi},$$ where the last step follows from integration by parts.

Answer 3

Let $I(a)$ be the integral

$$I(a)\equiv \int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx$$

Taking a derivative with respect to $a$ reveals that

$$\begin{align} I'(a)&=\frac{d}{da}\int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx\\\\ &=2a\int_0^{\infty}\frac{dx}{x^2+a^2}\\\\ &=\pi \end{align}$$

Integrating shows that $I(a)=\pi a +C$. Now, inasmuch as $I(0)=0$, $C=0$ and we find

$$\bbox[5px,border:2px solid #C0A000]{\int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx=\pi a}$$