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Multiple logarithmic integral

Let \zeta denote the Riemann zeta function. Evaluate the integral

    \[\mathcal{J} = \int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln \prod_{k=1}^{n} x_k \ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )\]

Solution

Based on symmetries,

    \begin{align*} \mathcal{J} &=n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln x_{1}\ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right ) \\ &=-n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1}\ln x_{1}\sum_{i=1}^{ \infty}\frac{(x_{1}\dots x_{n})^i}{i}\, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )\\ &=-n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n-1}}\int_{0}^{1}x_{1}^i\ln x_{1}dx_{1} \\ &=n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}} \end{align*}

Let \displaystyle \mathcal{S}_n = \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}}. It follows that

    \begin{align*} \mathcal{S}_n &= \sum_{i=1}^{\infty} \frac{1}{(i+1)^n} \left ( \frac{1}{i} - \frac{1}{i+1} \right ) \\ &= \mathcal{S}_{n-1} - \sum_{i=2}^{\infty} \frac{1}{i^{n+1}} \\ &= \mathcal{S}_{n-1} + 1 - \zeta (n +1) \end{align*}

Using the recursion we get that

    \[\mathcal{S}_n = n + 1 - \sum_{k=2}^{n+1} \zeta(k)\]

Thus,

    \[\mathcal{J} = n \left( n +1 - \sum_{k=2}^{n+1} \zeta(k) \right)\]

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An integral equality

Let \psi^{(1)} denote the trigamma function and let f be integrable on (0, 1). It holds that

    \[\int_{0}^{1} f \left ( \left \{ \frac{1}{x} \right \} \right )\, \mathrm{d}x = \int_{0}^{1} f(x) \psi^{(1)} (1+x) \; \mathrm{d}x\]

Solution

We have successively:

    \begin{align*} \int_{0}^{1} f \left(\left\{\frac{1}{x}\right\}\right) \mathrm{d}x &= \sum_{k=1}^{\infty}\int_{1/(k+1)}^{1/k} f \left(\left\{ \frac{1}{x} \right\} \right) \mathrm{d}x \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left( \left\{ u \right\} \right) \, \frac{\mathrm{d} u}{u^{2}} \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f\left( u-k \right) \, \frac{\mathrm{d} u}{u^{2}} \\ &= \sum_{k=1}^{\infty} \int_{0}^{1} f\left( v \right) \: \frac{\mathrm{d} v}{(v+k)^{2}} \\ &= \int_{0}^{1} f\left( v \right) \sum_{k=1}^{\infty} \frac{1}{(v+k)^{2}} \: \mathrm{d} v \\ &= \int_{0}^{1} f(v) \psi^{(1)}(v+1) \: \mathrm{d}v \end{align*}

where the interchange between the infinite sum and the integration is allowed by the uniform bound

    \[\left|\sum_{k=1}^{N} \frac{1}{(v+k)^{2}} \right| < \frac{\pi^{2}}{6} \quad , \quad N \geq 1, \, 0 \leq v \leq 1\]

The result follows.

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An series related to the eta Dedekind function

Prove that

    \[\sum_{n=1}^{\infty} \frac{1}{\sinh^2 n \pi} = \frac{1}{6} - \frac{1}{2\pi }\]

Solution

Let us consider the function

    \[f(z) = \frac{\cot \pi  z}{\sinh^2 \pi z}\]

and integrate it along a quadratic counterclockwise contour \Gamma_N with vertices \displaystyle \frac{N}{2} \left ( \pm 1 \pm i \right ) where N is a big odd natural number. Hence,

    \begin{align*} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \int_{N/2}^{-N/2} f \left ( x + \frac{i N}{2} \right )\, \mathrm{d}x + i \int_{N/2}^{-N/2} f \left ( i y + \frac{N}{2} \right ) \, \mathrm{d}y \\ &\quad \quad \quad + \int_{-N/2}^{N/2} f \left ( x - \frac{i N}{2} \right ) \, \mathrm{d}x + i \int_{-N/2}^{N/2} f \left ( iy + \frac{N}{2} \right ) \, \mathrm{d}y \end{align*}

We note that \displaystyle \lim_{N \rightarrow +\infty} f \left (x \pm \frac{i N}{2} \right ) = \frac{\mp i}{\cosh^2 \pi x} and that \displaystyle \lim_{N \rightarrow +\infty} f \left (i y \pm \frac{N}{2} \right ) = 0.

It’s also easy to see that

    \[\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2 \pi x } = \left [ \frac{\tanh \pi x}{\pi} \right ]_{-\infty}^{\infty} = \frac{2}{\pi}\]

Hence, as N \rightarrow +\infty we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = - \frac{4 i}{\pi}\]

By Residue theorem we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res} \left ( f; z = k \right ) + \mathfrak{Res} \left ( f; z = ik \right )\]

It is straightforward to show that

    \[\begin{matrix} \mathfrak{Res} \left ( f;z=k \right ) = &\mathfrak{Res} \left ( f;z=ik \right ) = & \dfrac{1}{\pi \sinh^2 \pi k} \\\\ \mathfrak{Res} \left ( f;z=0 \right ) & = & -\dfrac{2}{3\pi} \end{matrix}\]

Hence,

    \[\oint \limits_{\Gamma_N} f(z)\, \mathrm{d}z = 8 i\sum_{k=1}^{\infty}\frac{1}{\sinh^2 n \pi}-\frac{4 i}{3}\]

in the limit N \rightarrow +\infty. The result follows.

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Bessel function integral

Let J_0 denote the Bessel function of the first kind. Prove that

    \[\int_0^\infty J_0(x) \, \mathrm{d}x=1\]

Solution

We recall that

    \[J_0(x)=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}\]

Hence,

    \begin{align*} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}\\ &=\frac{1}{p}\sum_{n=0}^{\infty} \binom{-1/2}{n}\frac{1}{p^{2n}} \\ &= \frac{1}{\sqrt{1+p^2}} \end{align*}

Then,

    \[\lim_{p \rightarrow 0^+} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x = \lim_{p \rightarrow 0^+} \frac{1}{\sqrt{1+p^2}} =1\]

Using the fact that the J_0(x) looks like an ‘almost periodic’ function with decreasing amplitude. If we denote by \{\alpha_k\}_{k \geq 0} the zeros of J_0 then \alpha_k \nearrow \infty as k \to \infty and furthermore

    \[\left| \int_{\alpha_k}^{\infty} J_0(x)e^{-px}\,\mathrm{d} x \right| \leq \int_{\alpha_k}^{\alpha_{k+1}} |J_0(x)|e^{-px}\,\mathrm{d} x \rightarrow 0\]

as k \rightarrow \infty for each p \geq 0. So the integral converges uniformly in this case justifying the interchange of limit and integral.

The result follows.

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On multiplicative functions

Let \mu denote the Möbius function and \varphi the Euler’s totient function. Prove that

    \[\frac{n}{\varphi(n)}=\sum_{d \mid n} \frac{\mu^2(d)}{\varphi(d)}\]

Solution

Since both functions are multiplicative it suffices to prove the identity for prime numbers. Hence for n=p^k we have

    \begin{align*} \sum_{d \mid p^k} \frac{\mu^2(d)}{\varphi(d)} &=1+\frac{1}{\varphi(p)} \\ &=\frac{p}{p-1} \\ &=\frac{p^k}{p^{k-1}(p-1)} \\ &=\frac{p^k}{\varphi(p^k)} \end{align*}

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