3
$\begingroup$

I have to prove that $$ I=\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2} $$

I know that

$$ I_{+}=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{8}\log(2) $$ and I can write $$ I= \int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx+\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx $$

So I have only to evaluate $$I_{-}=\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx $$ Using the expansion of the logarithm valid in the range of integration I can also write $$I_{-}=-\sum_{n\geq1}\frac{1}{n}\int_{0}^{1}\frac{x^n}{1+x^2}dx $$ but then I can go any further because I can't find a simple solution that depend only on $n$ for the rational (and solvable) integral under the sum sign .

Maybe is simple to considerate instead of $ I_{-} $ the integral $\int_{0}^{1}\frac{\log(1-x^2)}{1+x^2}dx=I$ because it seems to me that $\int_{0}^{1}\frac{x^{2n}}{1+x^2}dx= \sum_{k=0}^{n-1} (\frac{(-1)^{n+k}}{2k+1} + \frac{\pi}{4}(-1)^n)$ but again I can go further.

Any help would be appreciated. Thanks

$\endgroup$
3
  • 2
    $\begingroup$ It might help you: $$\frac{x^n}{x^2+1}=\frac{x^n+x^{n-2}-x^{n-2}}{x^2+1}=x^{n-2}-\frac{x^{n-2}}{x^2+1}$$ With this, you can reduce all of the integrands either to $\frac{x}{x^2+1}$ or to $\frac{1}{x^2+1}$ plus some polynomials. $\endgroup$
    – Botond
    Aug 8, 2019 at 14:08
  • $\begingroup$ BTW, the sum is Catalan's constant. $\endgroup$
    – J.G.
    Aug 8, 2019 at 14:12
  • 1
    $\begingroup$ You should have: $$\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{4}\ln 2\color{red}-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$$ $\endgroup$
    – Zacky
    Aug 8, 2019 at 14:23

4 Answers 4

7
$\begingroup$

$$I_+=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx=\frac{\pi}{8}\ln 2\ ;\quad I_-=\int_0^1 \frac{\ln(1-x)}{1+x^2}dx$$ $$I_--I_+=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx\overset{(1-x)/(1+x)=t}=\int_0^1 \frac{\ln t}{1+t^2}dt$$ $$=\sum_{n=0}^\infty (-1)^n \int_0^1 t^{2n}\ln tdt=\sum_{n=0}^\infty \frac{(-1)^{n+\color{red}{1}}}{(2n+1)^2}=-G$$ $$\Rightarrow I_-=I_++I_--I_+=\frac{\pi}{8}\ln 2-G$$ Where $G$ is Catalan's constant.

$\endgroup$
4
  • $\begingroup$ How did you know that $$\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx\overset{(1-x)/(1+x)=t}=\int_0^1 \frac{\ln t}{1+t^2}dt=\sum_{n=0}^\infty (-1)^n \int_0^1 t^{2n}\ln t~dt$$ For the first equality, substituting $t=\frac{1-x}{1+x} \implies x = \frac{1-t}{t+1} \implies dx = -\frac{2}{(1+t)^2}~dt$ forms $$\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx=\int_{1}^0 \frac{\ln(t)}{1+\Big(\frac{1-t}{1+t}\Big)^2}\Big(\frac{-2}{(1+t)^2}\Big)dt$$ Maybe I did $dx$ wrong, I don't see how you found the second equality. I'm not sure what you did in the third equality. $\endgroup$
    – Axion004
    Aug 8, 2019 at 15:08
  • 1
    $\begingroup$ For the second equality note that the minus sign that you got there changes the bounds: $\int_a^b f(x)dx=-\int_b^a f(x) dx$. Also: $$\frac{1}{\frac{(1+t)^2+(1-t)^2}{(1+t)^2}}\frac{2}{(1+t)^2} =\frac{1}{1+t^2}$$ For the third equality just use the geometric series: $$\sum_{n=0}^\infty x^n= \frac{1}{1-x}\Rightarrow \sum_{n=0}^\infty (-x^2)^{n}=\frac{1}{1+x^2}$$ $\endgroup$
    – Zacky
    Aug 8, 2019 at 15:14
  • $\begingroup$ Thanks, I understand those two equalities. Then, at the last step, you have that $$\int_0^1 t^{2n}\ln tdt=\frac{-1}{(2n+1)^2}$$ which follows from integration by parts (unless there is an easier way). After that, you get Catalan's constant. $\endgroup$
    – Axion004
    Aug 8, 2019 at 15:46
  • 2
    $\begingroup$ Integration by parts is a way. However I do as follows: $$\int_0^1 x^q dx=\frac{1}{q+1} $$ $$\Rightarrow \frac{d}{dq} \int_0^1 x^q dx=\int_0^1 x^q \ln x dx= \frac{d}{dq} \frac{1}{q+1}=-\frac{1}{(q+1)^2}$$ And now set $q=2n$. $\endgroup$
    – Zacky
    Aug 8, 2019 at 15:51
2
$\begingroup$

\begin{align} I&=\int_{-1}^1\frac{\ln(1+x)}{1+x^2}\ dx\overset{x=\frac{1-y}{1+y}}{=}\int_0^\infty\frac{\ln2-\ln(1+y)}{1+y^2}\ dy\\ &=\frac{\pi}{2}\ln2-\int_0^\infty\frac{\ln(1+y)}{1+y^2}\ dy\\ &=\frac{\pi}{2}\ln2-\left(\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy+\underbrace{\int_1^\infty\frac{\ln(1+y)}{1+y^2}\ dy}_{\large y\mapsto 1/y}\right)\\ &=\frac{\pi}{2}\ln2-\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy-\int_0^1\frac{\ln(1+y)-\ln y}{1+y^2}\ dy\\ &=\frac{\pi}{2}\ln2-2\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy+\int_0^1\frac{\ln y}{1+y^2}\ dy\\ &=\frac{\pi}{2}\ln2-2\left(\frac{\pi}{8}\ln2\right)-G\\ &=\frac{\pi}{4}\ln2-G \end{align}

.


\begin{align} \int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&\overset{y=\frac{1-u}{1+u}}{=}\int_0^1\frac{\ln2-\ln(1+u)}{1+u^2}\ du\\ &=\frac{\pi}{4}\ln2-\int_0^1\frac{\ln(1+u)}{1+u^2}\ du\\ 2\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&=\frac{\pi}{4}\ln2\\ \int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&=\frac{\pi}{8}\ln2 \end{align}

$\endgroup$
1
$\begingroup$

Different method using harmonic series

\begin{align} \int_0^1\frac{\ln(1-x)}{1+x^2}\ dx&=\sum_{n=0}^\infty (-1)^n\int_0^1x^{2n}\ln(1-x)\ dx=-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{2n+1}\\ &=-\Im\sum_{n=1}^\infty\frac{i^n H_n}{n}=-\Im\left(\frac12\ln^2(1-i)+\operatorname{Li}_2(i)\right)=\frac{\pi}{8}\ln2-G \end{align}

$\endgroup$
1
$\begingroup$

Yet another way is to use Feynman's trick of differentiating under the integral sign. Let $$I(a) = \int_{-1}^1 \frac{\ln (1 + ax)}{1 + x^2} \, dx.$$ Note that $I(0) = 0$ and we require $I(1)$. Now \begin{align} I'(a) &= \int_{-1}^1 \frac{x}{(1 + ax)(1 + x^2)} \, dx\\ &= \frac{a}{1 + a^2} \int_{-1}^1 \frac{dx}{1 + x^2} + \frac{1}{1 + a^2} \int_{-1}^1 \frac{x}{1 + x^2} \, dx - \frac{a}{1 - a^2} \int_{-1}^1 \frac{dx}{1 + ax}\\ &= \frac{\pi a}{2(1 + a^2)} + \frac{1}{1 + a^2} \ln \left (\frac{1 - a}{1 + a} \right ) \end{align} Thus \begin{align} I(1) &= \frac{\pi}{4} \int_0^1 \frac{2a}{1 + a^2} \, da + \underbrace{\int_0^1 \ln \left (\frac{1 - a}{1 + a} \right ) \frac{da}{1 + a^2}}_{a \, \mapsto \, (1 - a)/(1 + a)}\\ &= \frac{\pi}{4} \ln 2 + \int_0^1 \frac{\ln a}{1 + a^2} \, da\\ &= \frac{\pi}{4} \ln 2 + \sum_{n = 0}^\infty (-1)^n \int_0^1 a^{2n} \ln x \, da\\ &= \frac{\pi}{4} \ln 2 + \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\int_0^1 a^{2n + s} \, da \right ]_{s = 0}\\ &= \frac{\pi}{4} \ln 2 + \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{2n + s + 1} \right ]_{s = 0}\\ &= \frac{\pi}{4} \ln 2 - \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^2}\\ &= \frac{\pi}{4} \ln 2 - \mathbf{G} \end{align}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .