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Bessel function integral

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

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


We recall that

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


    \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*}


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

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

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


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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Divergent Möbius series

Let \mu denote the Möbius function. Prove that

    \[\lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \frac{\left| \mu (n) \right|}{n} = +\infty\]


Summing only over primes , where |\mu(p)|=1 ,  we have that

    \[\lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \frac{\left| \mu (n) \right|}{n} \geq \sum_{p \in \mathcal{P}} \frac{1}{p} = +\infty\]

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Application of extended binomial theorem

For the values of a for which the following series makes sense, prove that

    \[\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} = \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\Gamma\left ( 2a+1 \right )}\]


We have successively:

    \begin{align*} \sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} &= \sum_{n=-\infty}^{\infty} \frac{\cos nx}{\left ( a+n \right )! \left ( a-n \right )!} \\ &=\frac{1}{\left (2a \right )!}\sum_{n=-\infty}^{\infty} \binom{2a}{a+n} \cos nx \\ &=\frac{1}{\left (2a \right )!}\mathfrak{Re} \left ( \sum_{n=-\infty}^{\infty} \binom{2a}{a+n} e^{inx} \right ) \\ &= \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\left ( 2a \right )!} \\ &= \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\Gamma\left ( 2a+1 \right )} \end{align*}

due to the extended binomial theorem.

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Double summation

Evaluate the sum

    \[\mathcal{S} = \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-j^2}{\left ( n^2+j^2 \right )^2}\]


Lemma: It holds that

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

Proof: Consider the function \displaystyle  f(z)=\frac{\cot \pi z}{\sinh^2 \pi z} and let us it integrate over the following contour \gamma

By the residue theorem it follows that

    \[\oint \limits_{\gamma} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res}_{z_k} f(z)\]

For the residues we have

    \[\begin{matrix} \displaystyle \mathfrak{Res}_{z=n}{\frac{\cot\pi z}{\sinh^2\pi z}} &= & \displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=ni}{\frac{\cot\pi z}{\sinh^2\pi z}} &= &\displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=0}{\frac{\cot\pi z}{\sinh^2\pi z}} & = &\displaystyle -\frac{2}{3\pi} \end{matrix}\]

The integrals along the sides vanish; hence:

    \[-2i\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=2\pi i\left(-\frac{2}{3\pi}+\frac{4}{\pi}\sum_{n=1}^\infty\frac{1}{\sinh^2\pi n}\right)\]

and since

    \[\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=\left[ \frac{1}{\pi}\tanh\pi x \right]_{-\infty}^\infty=\frac{2}{\pi}\]

the result follows. \blacksquare

Back to the problem. We have successively:

    \begin{align*} \mathcal{S} &= \frac{1}{2} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2}\sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \prod_{n=1}^{\infty} \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \frac{\sinh \pi j }{\pi j} \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \left ( \frac{1}{j^2} - \frac{\pi^2}{\sinh^2 \pi j} \right ) \\ &= \frac{\pi}{4} \end{align*}


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