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sinc/exp integral with a strange answer

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sinc/exp integral with a strange answer

Tue Aug 02, 2016 11:12 am

Posts: 38
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$$\int_{0}^{\infty}\frac{1}{xe^{x}}+\frac{\operatorname{sinc}(x)}{1-e^{x}}\,\text{d}x=\arg(i!)\approx-0.30164032$$

Me and LukeSQBacon have a solution, I'll post it later

Re: sinc/exp integral with a strange answer

Thu Aug 04, 2016 5:27 am

Posts: 30
Location: India

We recall the integral representation of Euler–Mascheroni constant $\displaystyle \int_0^{\infty} \left(\frac{1}{e^x-1} - \frac{1}{xe^{x}}\right)\,dx = \gamma$ and using $\displaystyle \int_0^{\infty} \frac{\sin x}{x}e^{-nx}\,dx = \arctan \frac{1}{n}$

\begin{align*}\int_0^{\infty} \frac{1}{xe^x}-\frac{\sin x}{x(e^x - 1)}\,dx &= \int_0^{\infty} \frac{1}{xe^x}- \frac{1}{e^x-1}\,dx + \int_0^{\infty} \frac{1}{e^x - 1} - \frac{\sin x}{x(e^x - 1)}\,dx \\&=-\gamma + \sum\limits_{n=1}^{\infty} \int_0^{\infty} \left(1-\frac{\sin x}{x}\right)e^{-nx}\,dx\\&= -\gamma + \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \arctan \frac{1}{n}\right) \\&= \Im \log \left[ e^{-\gamma i}\prod\limits_{n=1}^{\infty} \left(1+\frac{i}{n}\right)^{-1}e^{i/n}\right]\\&= \Im \log (\Gamma (1+i)) = \arg (i!)\end{align*}

where, we used the Weierstrass product formula of Gamma function $\displaystyle \Gamma(1+z) = e^{-\gamma z}\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{n}\right)^{-1}e^{\frac{z}{n}}$.

Re: sinc/exp integral with a strange answer

Thu Aug 04, 2016 1:08 pm

Posts: 38
Interesting solution! Quite convenient that the arctan sum can be rewritten in exactly the form of a Weierstrass factorization. Here is our solution, which is completely different to yours:

We first expand $\operatorname{sinc}(x)$ as a Taylor series around $x=0$ to get
\begin{align*} I&=\int_{0}^{\infty}\frac{1}{xe^{x}}+\frac{\operatorname{sinc}(x)}{1-e^{x}}\text{d}x \\ &=\int_{0}^{\infty}\frac{1}{xe^{x}}-\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n+1)!(e^{x}-1)}\text{d}x \\ &=\int_{0}^{\infty}\frac{1}{xe^{x}}-\frac{1}{e^{x}-1}-\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n+1)!(e^{x}-1)}\text{d}x \\ &=-\gamma-\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{(2n+1)!}\int_{0}^{\infty}\frac{x^{2n}}{e^{x}-1}\text{d}x\right) \\ &=-\gamma-\sum_{n=1}^{\infty}\frac{(-1)^{n}\Gamma(2n+1)\zeta(2n+1)}{(2n+1)!} \\ &=-\gamma-\sum_{n=1}^{\infty}\frac{(-1)^{n}\zeta(2n+1)}{2n+1} \end{align*}
Now use the series representation of the digamma function,
$$\psi(1+x)=-\gamma+\sum_{n=2}^{\infty}(-1)^{n}\zeta(n)x^{n-1}$$
Integrate it,
$$\log\Gamma(1+x)=-\gamma x+\sum_{n=2}^{\infty}\frac{(-1)^{n}\zeta(n)}{n}x^{n}$$
Substitute $x\mapsto -x$ and subtract the two series
$$\log\frac{\Gamma(1+x)}{\Gamma(1-x)}=-2\gamma x-2\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{2n+1}x^{2n+1}$$
Now substitute $x=i$ to get
\begin{align*} -\frac{i}{2}\log\frac{\Gamma(1+i)}{\Gamma(1-i)}&=-\gamma-\sum_{n=1}^{\infty}\frac{(-1)^{n}\zeta(2n+1)}{2n+1} \\ &=I \end{align*}
Then since $\overline{\Gamma(x+iy)}=\Gamma(x-iy)$, we can rewrite this as
\begin{align*} I&=-\frac{i}{2}\log\frac{\Gamma(1+i)}{\Gamma(1-i)} \\ &=\frac{1}{2}\Im\left(\log\frac{\Gamma(1+i)}{\Gamma(1-i)}\right) \\ &=\Im(\log\Gamma(1+i)) \\ &=\arg(i!) \end{align*}