# Evaluation of an improper integral

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.

• It is easy to find an antiderivative of that function, then find the limits at $0$ and $+\infty$. – enzotib Dec 13 '14 at 19:46
• When in doubt, part it out! – David H Dec 13 '14 at 19:52

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 $$\int_0^{+\infty} \log \left(1+\frac{1}{t^2}\right) \:\rm dt=\pi.$$

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• @N3buchadnezzar Yo can delete your email adress now. Thanks. – Olivier Oloa Dec 14 '14 at 13:48

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

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