MHB A fractional logarithm integral

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The integral $$\int_{0}^{1}\frac{\log(1+x^{2})}{1+x}dx$$ is proven to equal $$\frac{3}{4}\log^{2}(2)-\frac{\pi^{2}}{48}$$ through a series of transformations and the use of the logarithmic series. The discussion highlights the application of harmonic numbers and properties of the dilogarithm function, specifically $$\text{Li}_2(z)$$, to derive the result. Key steps involve simplifying alternating harmonic sums and utilizing relationships between dilogarithm values. The final expression confirms the equivalence to the desired result, demonstrating the effectiveness of complex analysis techniques in evaluating the integral. The proof is validated through a series of mathematical identities and transformations.
alyafey22
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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 ... :)
 
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ZaidAlyafey said:
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

$\chi$ $\sigma$
 
Hey chisigma , I love your solution . I got a question on how you simplified the alternating harmonic sum on the last step ?
 
$$I(a) = \int^1_0 \frac{\log(1+ax)}{1+x}\, dx $$

$$
\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.
 
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.
 
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