6

I was scrolling on the forum and I saw somebody solving the following integral :

$$\displaystyle\int_0^1 \frac{\ln(1+x)}{1+x^2} \mathrm{d}x$$

He used the really smart substitution $x= \frac{1-t}{1+t}$, but I have no idea how he thought about that.

It reminds me slightly the half angle tangent substitution (Weierstrass substitution ?).

So my questions are :

  1. Is it a well-known substitution ?
  2. When do we usually use it ? Are there indicators ?
  3. Can you give me an example where it works well ?

Thank you.

冥王 Hades
  • 3,083
LexLarn
  • 673

2 Answers2

5

It is well-known. Here are some noticeable examples of the substitution $x=\frac{1-t}{1+t}$

\begin{align} &\int_0^1 \frac{\ln\left(\frac1x-1\right)}{1+x}dx =\frac12\int_0^1 \frac{\ln2}{1+t}dt=\frac12\ln^22\\ \\ &\int_0^1 \frac1{1+x^3}dx =\int_0^1\frac{t+1}{3t^2+1}dt=\frac13\left(\ln2+\frac\pi{\sqrt3}\right)\\ \\ &\int_0^{\pi/4}\tanh^{-1}(\tan x)\ dx\overset{\tan x=\frac{1-t}{1+t}}= \frac12\int_0^1 \frac{\ln t}{1+t^2}dt=\frac12G \\ \end{align}

Quanto
  • 97,352
0

It is better (or it is, in my opinion at least) to use trigonometry here. If we let $x=\tan \theta$, then if $I=\int_{0}^{1}\frac{1}{1+x^2}\log(1+x) dx$ then $I=\int_{0}^{\pi/4}\log (1+\tan \theta) d\theta$.

Thus (if $u=\frac{\pi}{4}-\theta$), then $I=\int_{0}^{\pi/4}\log (1+\frac{1-\tan u}{1+\tan u}) du$. Hence $I=\frac{\pi}{4}\log 2-\int_{0}^{\pi/4}\log (1+\tan u) du$. So $I=-I+\frac{\pi}{4}\log 2$.

I think this question is from the STEP math exam.

Surajit131
  • 398