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Show that $\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx=\frac{\pi}8\ln 2$ using the change of variable $x=\tan y$.

Hint: $1+\tan x=\sqrt 2\sin(x+\pi/4)/\cos x$.

With the change of variable suggested I get

$$\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx=\int_0^{\pi/4}\ln(1+\tan y)\mathrm dy$$

but I dont know exactly what to do here or what to do with the identity $1+\tan x=\sqrt 2\sin(x+\pi/4)/\cos x$.

I tried some changes of variable or integration by parts but nothing work. I tried to write the logarithm as a series but the elements $\tan^k x$ are complicate to integrate. Some help will be appreciated, thank you.

Community
  • 1
Masacroso
  • 30,417

5 Answers5

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I suppose what you could do is write $$\log (1 + \tan y) = \frac{1}{2} \log 2 + \log \sin (y + \tfrac{\pi}{4} ) - \log \cos y,$$ then transform the second term with $v = \pi/4 - y$ to obtain $$\int_{y=0}^{\pi/4} \log \sin (y + \tfrac{\pi}{4}) \, dy = \int_{v=\pi/4}^0 \log \sin (\tfrac{\pi}{2} - v) \, (-dv) = \int_{v=0}^{\pi/4} \log \cos v \, dv.$$ Then this cancels with the integral of the third term, and you are left with $$\frac{\pi}{8} \log 2$$ as claimed.

heropup
  • 135,869
8

Here's a different solution.

Let

$$f=\int_0^1 \frac{\log (x+1)}{x^2+1} \, dx$$

Replacing the $\log$ by the defining integral in the form

$$\log (x+1)=\int_0^1 \frac{x}{x y+1} \, dy$$

we find a double integral form of $f$

$$f = \int _0^1\int _0^1\frac{x}{\left(x^2+1\right) (x y+1)}dydx$$

Exchanging now the order of integration, i.e. carrying out the x-integration first, which is elementary, gives

$$\int_0^1 \frac{x}{\left(x^2+1\right) (x y+1)} \, dx=\frac{\pi y}{4 y^2+4}+\frac{\log (4)}{4 y^2+4}-\frac{\log (y+1)}{y^2+1}$$

The integral over y now gives us the negative of the original integral f and two further terms which can be integrated to give

$$\int_0^1 \left(\frac{\pi y}{4 y^2+4}+\frac{\log (4)}{4 y^2+4}\right) \, dy=\frac{1}{4} \pi \log (2)$$

Hence we find that

$$f=\frac{1}{4} \pi \log (2)-f$$

or

$$f=\frac{1}{8} \pi \log (2)$$

QED.

Dr. Wolfgang Hintze
  • 12,465
4

Without using the given hint use the substitution $u = \frac{\pi}{4} -x$ to get $$I = \int_0^{\pi/4} \log (1 + \tan x) \, \mathrm{d}x = \int_0^{\pi/4} \log \left(\frac{2}{1 + \tan u}\right) \, \mathrm{d}u = \frac{\pi}{4}\ln 2 - I$$ since $1 + \tan (\pi/4 - u) = 1 + \frac{1-\tan u}{1 + \tan u} = \frac{2}{1 + \tan u}$.

So $2I = \frac{\pi}{4}\ln 2 \iff I = \frac{\pi}{8}\ln 2$

Zain Patel
  • 16,802
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$\displaystyle J=\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx$

Perform the change of variable $y=\dfrac{1-x}{1+x}$,

$\begin{align}\displaystyle J&=\int_0^1 \dfrac{\ln\left(\tfrac{2}{1+x}\right)}{1+x^2}dx\\ &=\int_0^1 \dfrac{\ln 2}{1+x^2}dx-J\\ &=\dfrac{\pi \ln 2}{4}-J \end{align}$

Therefore,

$2J=\dfrac{\pi \ln 2}{4}$

$\boxed{J=\dfrac{\pi \ln 2}{8}}$

FDP
  • 13,647
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Ok, I solved it, sorry :p.

We can observe that $\sin(x+\pi/4)$ is symmetric to $\cos x$ in $[0,\pi/4]$, so we can write

$$\begin{align}\int_0^{\pi/4}\ln(\sqrt 2\sin(x+\pi/4)/\cos x)\mathrm dx&=\int_0^{\pi/4}\frac12\ln 2\mathrm dx+\int_{\pi/4}^{\pi/2}\ln\sin x\mathrm dx-\int_0^{\pi/4}\ln\cos x\mathrm dx\\&=\frac{\ln2}2\int_0^{\pi/4}\mathrm dx=\frac{\pi}8\ln 2\end{align}$$

Masacroso
  • 30,417