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 Post subject: A beautiful log integralPosted: Sat Sep 10, 2016 12:45 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Prove that:

$$\int_{0}^{1}\frac{x \log x +1-x}{x \log^2 x} \log (1+x) \, {\rm d}x = \log \frac{4}{\pi}$$

Source

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 Post subject: Re: A beautiful log integralPosted: Thu Dec 08, 2016 5:07 pm

Joined: Tue May 10, 2016 3:56 pm
Posts: 33
First, for $p \geq 0$, we establish two integral formulas:
\begin{eqnarray}
\int_0^1\,\frac{x^p -1}{\ln x}\,dx & = &\ln(p+1).\\
\int_0^1\,\left(\frac{x^p -1}{x\ln^2 x} - \frac{p}{\ln x}\right)\,dx & =& p\ln p - p.
\end{eqnarray}
To this end, let $I(p) = \int_0^1\,\frac{x^p -1}{\ln x}\,dx$. Then
$$I'(p) = \int_0^1\,x^p\,dx = \frac{1}{p+1}.$$
In view of $I(0) = 0$, integrating this yields (1). Similarly, (2) follows from parametric differentiation and (1).

Denote the proposed integral as $T$. Using the power series of $\ln(1+x)$, we have
$$T = \sum_{n=1}^\infty\,(-1)^{n-1}\frac{1}{n}\,\int_0^1\,\frac{x\ln x+(1-x)}{x\ln^2x}x^n\,dx.$$
Rewrite
$$\frac{x\ln x+(1-x)}{x\ln^2x}x^n = \frac{x^n -1}{\ln x} + \left(\frac{x^n-1}{x\ln^2x} - \frac{n}{\ln x}\right) - \left(\frac{x^{n+1}-1}{x\ln^2x} - \frac{n+1}{\ln x}\right).$$
Appealing to the formulas (1) and (2) gives
\begin{eqnarray*}
T &= &\sum_{n=1}^\infty\,(-1)^{n-1}\frac{1}{n}\left(1 - n\ln\left(\frac{n+1}{n}\right)\right)\\
& = & \sum_{n=1}^\infty\,(-1)^{n-1}\left(\frac{1}{n}- \ln\left(\frac{n+1}{n}\right)\right)\\
& = & \ln \frac{4}{\pi}.
\end{eqnarray*}

Remark: It is interesting to see that
$$\sum_{n=1}^\infty\,\left(\frac{1}{n}- \ln\left(\frac{n+1}{n}\right)\right) = \gamma.$$

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 Post subject: Re: A beautiful log integralPosted: Fri Dec 09, 2016 7:52 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
mathofusva wrote:
Remark: It is interesting to see that
$$\sum_{n=1}^\infty\,\left(\frac{1}{n}- \ln\left(\frac{n+1}{n}\right)\right) = \gamma.$$

We have seen it in here .

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: A beautiful log integralPosted: Fri Dec 09, 2016 8:48 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
A similar integral to that of $(2)$ used by mathofusva is the following:

$$\int_0^1 \left( \frac{1}{1-x} + \frac{1}{\ln x} \right) \, {\rm d}x =\gamma$$

where $\gamma$ is the Euler - Mascheroni constant.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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