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While trying to find several references to answer Pranav's problem, I encounter the following multiple integrals

$$I=\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$$

Question :

Does the above integral have a closed form? If it has a closed form, how to obtain it?


According to Chriss'sis in the chat room, the integral has a nice closed form $$\frac{2\ln 2 \pi -2 \gamma -5}{4}$$ but no proof given so far. Since the term $\ln 2 \pi$ appears in the close form, I have a hunch that it would involve Stirling's approximation. After trying to evaluate it for hours, I think it is about time to ask it on the main page. Here is my attempt so far: \begin{align} I&=\int_0^1\int_0^1\left(1-x^y\right)\;\mathrm dy\int_0^1\left(1-x^z\right)\;\mathrm dz\frac{\ln x}{(1-x)^3}\,\mathrm dx\\[7pt] &=\int_0^1\frac{\ln x+1-x}{\ln x}\cdot\frac{\ln x+1-x}{\ln x}\cdot\frac{\ln x}{(1-x)^3}\,\mathrm dx\\[7pt] &=\int_0^1\frac{\left(\ln x+1-x\right)^2}{(1-x)^3\ln x}\,\mathrm dx\\ \end{align} I'm stuck in the latter expression. I'm thinking of the following parametric integral $$I(s)=\int_0^1\frac{x^s\left(\ln x+1-x\right)^2}{(1-x)^3\ln x}\,\mathrm dx\quad\implies\quad I'(s)=\int_0^1\frac{x^s\left(\ln x+1-x\right)^2}{(1-x)^3}\,\mathrm dx$$ then expanding the quadratic term and using series representation of $$\frac{1}{(1-x)^3}=\frac{1}{2}\sum_{n=1}^\infty n(n-1)x^{n-2}$$ but the calculation would be cumbersome. Is there a better way? Thanks in advance for your help.

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Continuing from Chris's sis answer, consider the partial sum:

$$\frac{1}{2}\sum_{n=1}^m n(n+1)\left(\ln(n)-2\ln(n+1)+\ln(n+2)\right)+1-\frac{1}{n}$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\sum_{n=1}^m\left( \ln\left(\frac{n}{n+1}\right)^{n(n+1)}+\ln\left(\frac{n+2}{n+1}\right)^{n(n+1)}\right)$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\left(\ln\left(\dfrac{\displaystyle \prod_{k=1}^m k^{2k}}{(m+1)^{m(m+1)}}\right)+\ln\left(\dfrac{(m+2)^{m(m+1)}}{\displaystyle \prod_{k=1}^m (k+1)^{2k}}\right)\right)$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\ln\left(\frac{(m+2)^{m(m+1)}}{(m+1)^{m(m+1)}}\cdot \frac{(m!)^2}{(m+1)^{2m}}\right)$$

Use Stirling's approximation and rewrite the expression as:

$$\frac{1}{2}\left(\ln(2\pi)-H_m+\ln m+\ln\left(\left(\frac{m+2}{m+1}\right)^{m(m+1)}\left(\frac{m}{m+1}\right)^{2m}\frac{1}{e^m}\right)\right)$$

I still need to evaluate the logarithmic limit by hand but wolfram alpha gives $-5/2$ i.e the final result is: $$\frac{1}{2}\left(\ln(2\pi)-\gamma-\frac{5}{2}\right)$$

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  • $\begingroup$ (+1) $\displaystyle \lim\limits_{m \to \infty} \ln\left(\frac{m}{m+1}\right)^{2m} = \ln e^{-2} = -2$ and $\lim\limits_{m \to \infty} m(m+1)\ln\left(1+\frac{1}{m+1}\right) - m = \lim\limits_{m \to \infty} -\frac{m(m+1)}{2(m+1)^2} + \mathcal{O}(1/m) = -1/2$ by taylor expansion of $ln$ :-) $\endgroup$ – r9m Dec 22 '14 at 2:56
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Using series representation you specified above, I got that your integral gets reduced to $$\sum _{n=1}^{\infty } \frac{(n+1) n^2 (\log (n)-2 \log (n+1)+\log (n+2))+n-1}{2 n}$$

Can you take it from here?

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    $\begingroup$ (+1) but it will be very helpful if you added the steps that lead to this summation form as well ! :-) Thanks :) $\endgroup$ – r9m Dec 22 '14 at 2:57
  • $\begingroup$ @OFFSHARING Did you change your username? $\endgroup$ – user285523 Nov 27 '15 at 18:06
  • $\begingroup$ @0.5772156649... Yes, my new username means that I totally ceased sharing math knowledge in any math community since now on. $\endgroup$ – user 1357113 Nov 27 '15 at 19:09

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