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How can the following integral be calculated: $$ I_n=\int_0^1\int_0^1\cdots\int_0^1\frac{\prod_{k=1}^{n}\left(\frac{1-x_k}{1+x_k}\right)}{1-\prod_{k=1}^{n}x_k}dx_1\cdots dx_{n-1}dx_n $$ There should be $n$ integral signs, but I didn't know how to write that.

It is easy to show that $I_1=\ln(2)$. After partial fractioning and the help of Wolfram Alpha, I managed to show that $I_2=4\ln(2)-2\ln^2(2)-\frac{\pi^2}{6}$.

But how to derive a general result? Any help would be highly appreciated!

Edit:

As a supplementary question, how to calculate this slightly modified integral: $$ J_n=\int_0^1\int_0^1\cdots\int_0^1\frac{\prod_{k=1}^{n}\left(\frac{1-x_k}{1+x_k}\right)}{1+\prod_{k=1}^{n}x_k}dx_1\cdots dx_{n-1}dx_n $$ Again, it can be shown easily, that $J_1=1-\ln(2)$.

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    $\begingroup$ FWIW $$I_3 =\pi ^2 \log (2) + 4 \log ^3(2) - \frac{27 \zeta (3)}{4}$$ $\endgroup$ Nov 13, 2014 at 9:09
  • $\begingroup$ @user109899 : Is there any particular reason why you are interested in this integral? As it seems $I_n$ will be related to $\zeta(n)$. Did you envisage to use this integral to investigate some properties of $\zeta(n)$? $\endgroup$
    – Przemo
    Jun 30, 2015 at 16:51

2 Answers 2

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In this answer, I will provide a method that uses iterated partial summation on the alternating Harmonic numbers that will allow you to explicitly evaluate either of these integrals for any $n$.

Start by expanding the series $$\frac{1}{1-\prod_{k}x_{k}}=1+\prod_{k}x_{k}+\prod_{k}x_{k}^{2}+\cdots$$ so that $$I_{n}=\sum_{j=0}^{\infty}\left(\int_{0}^{1}\frac{1-x}{1+x}x^{j}dx\right)^{n}=\sum_{j=0}^\infty r_j^n.$$ Next we will rewrite $r_j$ in a more manageable form. Expanding the power series we have that $$r_j=\int_{0}^{1}\frac{1-x}{1+x}x^{j}dx=\sum_{k=0}^{\infty}(-1)^{k}\int_{0}^{1}(1-x)x^{j+k}dx,$$ and by the Beta function identity $$\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ we obtain $$r_j=\sum_{k=0}^{\infty}(-1)^{k}\frac{1}{(k+j+1)(k+j+2)}.$$ The terms in this series may be split by partial fractions yielding $$(-1)^{j}\sum_{k=0}^{\infty}\frac{(-1)^{k+j}}{k+j+1}+(-1)^{j}\sum_{k=0}^{\infty}\frac{(-1)^{k+j+1}}{k+j+2}$$ and since $\sum_{k=0}^\infty (-1)^k/(k+1)=\log2$ we have $$r_j=(-1)^{j}\left(2\log2-H_j'-H_{j+1}'\right)$$ where $$H_j'=\sum_{k=1}^{j}\frac{(-1)^{k-1}}{k}$$ is the $j^{th}$ alternating harmonic number. To evaluate the series, we need only compute the partial sums of $$\sum_{j=0}^m (-1)^{nj}\left(2\log2-H_j'-H_{j+1}'\right)^n$$ for any $n,m$. This can be done explicitly using the technique of iterated partial summation as shown in this answer. This will yield an exact answer for any $n$.

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Since all the $a_k \lt 1$ (except at one point), I would try expanding $$\frac{1}{1-\prod_k^n a_k}=1+\left(\prod_k^n a_k \right)+\left(\prod_k^n a_k \right)^2+\cdots$$ (Geometric series) This will allow you to turn your problem into a sum of integrals in which, in every term of the sum, the $a_k$'s appear multiplicatively and thus can be integrated independently via Fubini:

$$\sum_{i=0}^\infty \int_0^1 \cdots \int_0^1 \prod_{k=1}^n \left( \frac{1-a_k}{1+a_k} (a_k)^i\right) $$. Which is equal to

$$\sum_{i=0}^\infty \left( \int_0^1 \left( \frac{1-a_k}{1+a_k} (a_k)^i\right) \right)^n $$

I can't assure this will work, but it seems to me that the problem has been reduced to computing one integral (which doesn't look impassable) and a series.

Hope this helps and let me know if it does.

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    $\begingroup$ Yes, I thought of that; the Integral is equal to $\psi\left(\frac{i+2}{2}\right)-\psi\left(\frac{i+1}{2}\right)-\frac{1}{i+1}$ and after that, it gets pretty complicated. But thank you for your answer! $\endgroup$ Nov 5, 2014 at 1:49
  • $\begingroup$ what is $\psi$? And why is it involving complex numbers? $\endgroup$
    – alonso s
    Nov 5, 2014 at 1:51
  • $\begingroup$ Sorry, $\psi$ denotes the digamma function, defined as $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ where $\Gamma$ denotes Euler's Gammafunction. There are no complex numbers involved, i just used the index $i$ of your sum to express the integral in terms of it. $\endgroup$ Nov 5, 2014 at 1:55

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