Evaluation of an improper integral

Question

I was trying earlier today to prove the convergence of an improper integral. In order to test my conclusion, I plugged it in WolframAlpha, getting a really beautiful answer:

$$ \int\limits_0^\infty\log\left(1+\frac{1}{t^2 }\right)\,\mathrm dt=\pi $$

I'm intersted if there's a way to prove this result other than the way Wolfram solves it through brute force.

Answer 1

Observe that, integrating by parts, we have $$ \int \log t \:\rm dt = t \log t - t $$ and $$ \int \log (1+t^2) \:\rm dt = t \log (1+t^2) - 2\int \frac{t^2}{1+t^2}\:\rm dt =t \log (1+t^2) -2 t+2 \arctan t $$ thus $$ \begin{align} \int \log \left(1+\frac{1}{t^2}\right) \:\rm dt&=\int \log (1+t^2) \:\rm dt -2\int \log t \:\rm dt\\\\ & =2 \arctan t+t\log \left(1+\frac{1}{t^2}\right) \end{align} $$ and, since $\displaystyle \lim_{t \rightarrow +\infty} \arctan t=\frac \pi 2 $ and $\displaystyle \lim_{t \rightarrow +\infty} t\log \left(1+\frac{1}{t^2}\right)=0 $ we get $$ \int_0^{+\infty} \log \left(1+\frac{1}{t^2}\right) \:\rm dt=\pi. $$

Answer 2

Wolfram Alpha gives the indefinite integral $\displaystyle\int \ln\left(1+\frac{1}{t^2}\right)\,\mathrm dt = t\ln\left(1+\frac{1}{t^2}\right) + 2\arctan t$.

It is easy to see that integration by parts will work.

Set $u = \ln\left(1+\dfrac{1}{t^2}\right)$ and $\mathrm dv =\mathrm dt$, we have $\mathrm du = \dfrac{1}{1+\frac{1}{t^2}} \cdot \dfrac{-2}{t^3}\,\mathrm dt = \dfrac{-2}{t^3+t}$ and $v = t$.

Thus, $\displaystyle\int \ln\left(1+\frac{1}{t^2}\right)\,dt = t\ln\left(1+\frac{1}{t^2}\right)- \int t \dfrac{-2}{t^3+t}\,\mathrm dt$ $= t\ln\left(1+\frac{1}{t^2}\right) + \int \dfrac{2}{t^2+1}\,\mathrm dt$ $= t\ln\left(1+\dfrac{1}{t^2}\right) + 2\arctan t$.

Since $\displaystyle\lim_{t\to 0^+}t\ln\left(1+\dfrac{1}{t^2}\right) + 2\arctan t = 0$ and $\displaystyle\lim_{t\to \infty}t\ln\left(1+\dfrac{1}{t^2}\right) + 2\arctan t = \pi$, we have

$\displaystyle\int_{0}^{\infty} \ln\left(1+\frac{1}{t^2}\right)\,\mathrm dt = \pi$.