While browsing on Integral and Series, I found a strange integral posted by @Sangchul Lee. His post doesn’t have a response for more than a month, so I decide to post it here. I hope he doesn’t mind because the integral looks very interesting to me. I hope for you too. 🙂

How does one prove∫∞−∞dx1+[x+tan(x)]2=π?

Please don’t ask me, I really have no idea how to prove it. I hope users here can find the answer to prove the integral. I’m also interested in knowing any references related to this integral. Thanks in advance.

**Answer**

Here is an approach.

We may use the following result, which goes back to G. Boole (1857) :

∫+∞−∞f(x−a1x−λ1−⋯−anx−λn)dx=∫+∞−∞f(x)dx

with ai>0,λi∈R and f sufficiently ‘regular’.

Observe that, for x≠nπ, n=0,±1,±2,…, we have cotx=limN→+∞(1x+1x+π+1x−π+⋯+1x+Nπ+1x−Nπ) leading to (see Theorem 10.3 p. 14 here and see achille’s answer giving a route to prove it)

∫+∞−∞f(x−cotx)dx=∫+∞−∞f(x)dx

with f(x)=11+(π2−x)2.

On the one hand, from (2),

∫+∞−∞f(x−cotx)dx=∫+∞−∞f(x)dx=∫+∞−∞11+(π2−x)2dx=∫+∞−∞11+x2dx=π

On the other hand, with the change of variable x→π2−x,

∫+∞−∞f(x−cotx)dx=∫+∞−∞f(π2−x−tanx)dx=∫+∞−∞11+(x+tanx)2dx

Combining (3) and (4) gives

\int_{-\infty}^{+\infty}\frac{1}{1+\left(x+ \tan x \right)^2} \mathrm{d}x=\pi.

**Attribution***Source : Link , Question Author : Venus , Answer Author : Community*