MHB A fractional logarithm integral

[alyafey22]

Prove the following

$$\int_{0}^{1}\frac{\log(1+x^{2})}{1+x}dx=\frac{3}{4}\log^{2}(2)-\frac{{\pi}^{2}}{48} $$

Good luck with this one ... :)

 

[chisigma]

Prove the following

$$\int_{0}^{1}\frac{\log(1+x^{2})}{1+x}dx=\frac{3}{4}\log^{2}(2)-\frac{{\pi}^{2}}{48} $$

Good luck with this one ... :)

Remember the well known logarithmic series we write...

$\displaystyle \frac{\ln (1+ x^{2})}{1+x} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\ \frac{x^{2 n}}{1+x}\ (1)$

... and now we remember the formula...

$\displaystyle \int_{0}^{1} \frac{x^{2 n}}{1 + x}\ dx = \frac{1}{2}\ (H_{n} - H_ {n-\frac{1}{2}})\ (2)$

... where $H_{n}$ is the Harmonic Number of order n, and the formula...

$\displaystyle \frac{H_{n} - H_ {n-\frac{1}{2}}}{2} = \ln 2 + \sum_{k=1}^{n} \frac{(-1)^{k}}{k}\ (3)$

... so that we obtain...

$\displaystyle \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x}\ dx = \ln 2\ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\ \sum_{k=1}^{n} \frac{(-1)^{k}}{k} =$

$\displaystyle = \ln^{2} 2 - \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} - \frac{1}{4}\ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}} = \frac{3}{4}\ \ln^{2} 2 - \frac{\pi^{2}}{48}\ (4)$

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

Hey chisigma , I love your solution . I got a question on how you simplified the alternating harmonic sum on the last step ?

 

[alyafey22]

$$I(a) = \int^1_0 \frac{\log(1+ax)}{1+x}\, dx $$

$$
\begin{align*}
I'(a) = \int^1_0 \frac{x}{(1+x)(1+ax)}\, dx &= \frac{1}{1-a} \left(\int^1_0\frac{1}{(1+ax)}\, dx -\int^1_0 \frac{1}{(1+x)}\right)\\
&= \frac{1}{1-a} \left(\frac{1}{a} \log(1+a)-\log(2) \right)\\
&= \frac{\log(1+a)}{a(1-a)}-\frac{\log(2)}{a-1} \\
&= \frac{\log(1+a)}{1-a}+\frac{\log(1+a)}{a} -\frac{\log(2)}{1-a} \\

\end{align*}
$$Using http://www.mathhelpboards.com/f10/generalized-fractional-logarithm-integral-5467/#post24937we obtain $$I(a)=-\text{Li}_2 \left(\frac{1}{2} \right) +\text{Li}_2 \left(\frac{1-a}{2}\right)-\text{Li}_2(-a)+C$$using $$I(0)=0$$ we obtain $$C=0$$$$I(a) = \int^1_0 \frac{\log(1+ax)}{1+x}\, dx =- \text{Li}_2 \left(\frac{1}{2} \right) +\text{Li}_2 \left(\frac{1-a}{2}\right) -\text{Li}_2(-a)\,\,\,\, \text{valid for }0\leq\text{Re}(a) <1$$Now we make a little trick $$I(i)+I(-i) = \int^1_0 \frac{\log(1+x^2)}{1+x}\, dx$$

$$ \int^1_0 \frac{\log(1+x^2)}{1+x}\, dx = -2\text{Li}_2 \left(\frac{1}{2} \right) +\text{Li}_2 \left(\frac{1-i}{2}\right)+ \text{Li}_2 \left(\frac{1+i}{2}\right) -\text{Li}_2(-i)-\text{Li}_2(i)$$

The result is numerically equivalent to the result we are looking for and it can be simplified , I know the answer looks nasty (Tmi) , but the complex conjugate that appears on the logarithms is rather promising ,furthermore this will allow us to generlaize the integral in this http://www.mathhelpboards.com/f10/generalized-fractional-logarithm-integral-5467/#post24937.

 

[alyafey22]

To complete the solution we can use the following

1-$$\operatorname{Li}_{\,n}(-z) + \operatorname{Li}_{\,n}(z) = 2^{1-n} \,\operatorname{Li}_{\,n}(z^2)$$

Hence

$$-\operatorname{Li}_{\,2}(-i) - \operatorname{Li}_{\,2}(i) =- \frac{1}{2} \,\operatorname{Li}_{\,2}(-1) = \frac{\pi^2}{24}$$

2-$$\operatorname{Li}_2(z) + \operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\,\,\,$$

$$\operatorname{Li}_2\left(\frac{1+i}{2}\right) + \operatorname{Li}_{2}\left(1-\frac{1+i}{2}\right) = \frac{\pi^2}{6}-\log\left(\frac{1+i}{2}\right) \log \left(1-\frac{1+i}{2} \right) \,\,\,\,=\frac{\pi^2}{6}-\frac{\log^2(2)}{4}-\frac{\pi^2}{16}$$

Hence we have

$$\int^1_0 \frac{\log(1+x^2)}{1+x}\, dx=-\frac{\pi^2}{6}+\log^2(2)+-\frac{\log^2(2)}{4}+\frac{5\pi^2}{48}+\frac{\pi^2}{24}=\frac{3}{4}\log^2(2)-\frac{\pi^2}{48}$$

I am using the principle logarithm ... For the proofs of the identities you can see my lessons on http://www.mathhelpboards.com/f10/advanced-integration-techniques-3233/index3.html.