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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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Fourier transformation

Let \xi \in \mathbb{R} and \alpha \in (0, 1). Show that

    \[\int_{- \infty}^{\infty} {e}^{-2 \pi i x \xi} \frac{ \sin \pi \alpha }{ \cosh \pi x + \cos \pi \alpha } \mathrm{d} x = \frac{2 \sinh 2 \pi \alpha \xi }{ \sinh 2 \pi \xi }\]

Solution

We note that

    \begin{align*} \frac{\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha} &= \frac{1}{1+e^{-\pi(x+i\alpha)}}-\frac{1}{1+e^{-\pi(x - i\alpha)}}\\ &= 2\sum\limits_{n=1}^{\infty} (-1)^{n}e^{-n\pi x}\sin n\pi\alpha \end{align*}

Thus,

    \begin{align*} \int_{-\infty}^{\infty} \frac{e^{-2\pi i \xi x}\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x &= \int_{0}^{\infty} \frac{2\sin \pi \alpha \cos 2\pi\xi x}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x\\ &= 4\int_0^{\infty}\sum\limits_{n=1}^{\infty}(-1)^n e^{-n\pi x}\cos 2\pi\xi x \sin n\pi \alpha\,\mathrm{d}x\\ &= \frac{4}{\pi}\sum\limits_{n=1}^{\infty}(-1)^n \frac{n\sin n\pi\alpha}{n^2+4\xi^2}\\ &= \frac{2}{\pi}\sum\limits_{n=1}^{\infty} (-1)^n\left(\frac{1}{n+2i\xi}+\frac{1}{n-2i\xi}\right)\sin n\pi \alpha\\ &= \frac{2}{\pi}\sum\limits_{n = -\infty}^{\infty} \frac{(-1)^n\sin n\pi\alpha}{n+2i\xi}\\ &= \frac{2 \sin 2\pi i \alpha\xi}{\sin 2\pi i\xi} \qquad \textrm{ (by residue theorem) }\\ &= \frac{2 \sinh 2\pi \alpha\xi}{\sinh 2\pi \xi} \end{align*}

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Arctan integral

Prove that

    \[\int_0^{\infty}\frac{\arctan x -x e^{-x}}{x^2}\,\mathrm{d}x=1+\gamma\]

where \gamma denotes the Euler – Mascheroni constant.

Solution

Beginning by parts we have,

    \begin{align*} \int_{0}^{\infty} \frac{\arctan x- xe^{-x}}{x^2} \, \mathrm{d}x &= \left [ -\frac{\arctan x - xe^{-x}}{x} \right ]_0^\infty + \\ &\quad \quad \quad + \int_{0}^{\infty} \frac{1}{x} \left ( \frac{1}{x^2+1} +e^{-x} (x-1) \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x\left ( x^2+1 \right )} + \frac{e^{-x}(x-1)}{x} \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x} - \frac{x}{x^2+1} + e^{-x} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x \end{align*}

However,

\begin{aligned} \int_{\epsilon}^{M} \left ( \frac{1}{x} - \frac{x}{x^2+1} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x &= \left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \int_{\epsilon}^{M} \frac{e^{-x}}{x} \, \mathrm{d}x \\ &=\left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \left [ \ln x e^{-x} \right ]_\epsilon^{M}  - \int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \\ &=\left ( \ln M - \frac{\ln \left ( M^2+1 \right )}{2} \right ) + \frac{\ln \left ( \epsilon^2+1 \right )}{2} - \\ &\quad \quad \quad \quad -\ln M e^{-M} + \left (\ln \epsilon \; e^{-\epsilon} - \ln \epsilon \right ) - \\ &\quad \quad \quad \quad  -\int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \end{aligned}

The result now follows taking \epsilon \rightarrow 0^+ and M \rightarrow +\infty.

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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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Trigonometric integral

Evaluate the integral

    \[\mathcal{J} =\int \frac{\mathrm{d}x}{\sin x \sin (x+1)}\]

Solution

We have successively:

    \begin{align*} \mathcal{J} &= \int \frac{\mathrm{d}x}{\sin x \sin (x+1)} \\ &= \frac{1}{\sin 1} \int \frac{\sin 1}{\sin x \sin (x+1)} \, \mathrm{d}x \\ &= \frac{1}{\sin 1} \int \frac{\sin \left ( x+1-x \right )}{\sin x \sin (x+1)}\, \mathrm{d}x \\ &= \frac{1}{\sin 1} \int \frac{\sin (x+1) \cos x - \cos (x+1) \sin x}{\sin x \sin (x+1)} \, \mathrm{d}x\\ &= \frac{1}{\sin 1} \int \left ( \frac{\sin (x+1) \cos x}{\sin x \sin (x+1)} - \frac{\cos (x+1) \sin x}{\sin x \sin (x+1)} \right ) \, \mathrm{d}x \\ &= \frac{1}{\sin 1} \int \left ( \frac{\cos x}{\sin x} - \frac{\cos (x+1)}{\sin (x+1)} \right ) \, \mathrm{d}x \\ &= \frac{1}{\sin 1} \left ( \ln \sin x - \ln \sin (x+1) \right ) + c \; , \; c \in \mathbb{R} \end{align*}

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