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.

  • $\begingroup$ It is easy to find an antiderivative of that function, then find the limits at $0$ and $+\infty$. $\endgroup$ – enzotib Dec 13 '14 at 19:46
  • $\begingroup$ When in doubt, part it out! $\endgroup$ – 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 $$ \int_0^{+\infty} \log \left(1+\frac{1}{t^2}\right) \:\rm dt=\pi. $$

  • $\begingroup$ @N3buchadnezzar You can give your email adress here, if you want to, then I will email you :). $\endgroup$ – Olivier Oloa Dec 14 '14 at 13:31
  • $\begingroup$ @N3buchadnezzar Yo can delete your email adress now. Thanks. $\endgroup$ – 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.