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For $n\in\mathbb{N}$, I'm trying to find a closed form for the following integrals : $$I_n=\int_{0}^{\infty}e^{-x}\sin(n\ln(x))\text{d}x$$

My real objective is to evaluate $\sum\limits_{n=1}^{\infty}\frac{I_n}{n}$, and since interchanging the sum and the integral didn't lead anywhere, I suppose that finding a closed form expression for $I_n$ is the way to go, but I'm lost as of how to proceed...

Maybe the residue theorem/contour integration could help, but I'm not familiar with complex analysis so I haven't tried it - feel free to use it though.

Any suggestion ?

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  • $\begingroup$ Mathematica gives the following result: $I_n = \frac{1}{2} i (\Gamma (1-i n)-\Gamma (i n+1))$. $\endgroup$
    – JimB
    Feb 4, 2019 at 0:14
  • $\begingroup$ A closed-form is many times in the eye of the beholder. The infinite sum appears close to -0.132333390711525914530. (In my previous comment which I deleted I forgot to divide $I_n$ by $n$ in the infinite sum.) $\endgroup$
    – JimB
    Feb 4, 2019 at 4:17

3 Answers 3

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By definition, for any $z\in \mathbb{C}$ for which $\Re{(z)}>-1$ : $$\Gamma(1+z):=\int_0^\infty x^z\cdot e^{-x}\text{d}x$$ and by the Euler identity $$\sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$ Hence $$I_n=\int_0^\infty e^{-x}\sin (n\ln x)=\int_0^\infty e^{-x}\frac{e^{in\ln x}-e^{-in\ln x}}{2i}=\int_0^\infty e^{-x}\frac{x^{in}-x^{-in}}{2i}=\frac{\Gamma(1+in)-\Gamma(1-in)}{2i}$$

Now using the familiar identity $z\cdot \Gamma(z)=\Gamma(z+1)$ we can proceed to $$\frac{I_n}{n}=\frac{\Gamma(1+in)-\Gamma(1-in)}{2in}=\frac{in\Gamma(in)+in\Gamma(-in)}{2in}=\frac{\Gamma(in)+\Gamma(-in)}{2}$$

So at least formally: $$\sum_{n=1}^\infty \frac{I_n}{n}=\frac{1}{2}\sum_{n=-\infty,n\not = 0}^\infty\Gamma(in)$$

On other-hand, since $\Gamma(\bar{z})=\overline{\Gamma(z)}$ we get $$\sum_{n=1}^\infty \frac{I_n}{n}=\sum_{n=1}^\infty \frac{\Gamma(in)+\Gamma(-in)}{2}=\sum_{n=1}^\infty \frac{\Gamma(in)+\overline{\Gamma(in)}}{2}=\sum_{n=1}^\infty \Re{(\Gamma(in))}$$

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  • $\begingroup$ Thank you very much, this is a nice representation ! $\endgroup$ Feb 5, 2019 at 20:51
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A rather nice expression for the sum can be found using the Fourier series $$ \sum \limits_{n=1}^\infty \frac{\sin(n y)}{n} = \frac{\pi}{2} \left[1 - 2 \left\{\frac{y}{2\pi}\right\} \right] \, , \, y \in \mathbb{R} \, , $$ where $\{z\} = z - \lfloor z \rfloor$ is the fractional part of $z \in \mathbb{R}$ . With this definition it satisfies $\{-z\} = 1- \{z\}$ for $z \in \mathbb{R} \setminus \mathbb{Z}$ .

We can write the sum as \begin{align} S &\equiv - \sum \limits_{n=1}^\infty \frac{I_n}{n} = \frac{\pi}{2} \int \limits_0^\infty \mathrm{e}^{-x} \left[2 \left\{\frac{\ln(x)}{2\pi}\right\} - 1\right] \, \mathrm{d} x = \frac{\pi}{2} \int \limits_{-\infty}^\infty \mathrm{e}^{-\mathrm{e}^{t}+t} \left[2 \left\{\frac{t}{2\pi}\right\} - 1\right] \, \mathrm{d} t\\ &= \frac{\pi}{2} \int \limits_0^\infty \left(\mathrm{e}^{-\mathrm{e}^{t}+t} \left[2 \left\{\frac{t}{2\pi}\right\} - 1\right] + \mathrm{e}^{-\mathrm{e}^{-t}-t} \left[2 \left(1-\left\{\frac{t}{2\pi}\right\}\right) - 1\right] \right)\, \mathrm{d} t \\ &= \frac{\pi}{2} \int \limits_0^\infty \left(\mathrm{e}^{-\mathrm{e}^{-t}-t} - \mathrm{e}^{-\mathrm{e}^{t}+t} \right) \left(1 - 2\left\{\frac{t}{2\pi}\right\} \right) \, \mathrm{d}t \\ &= \frac{\pi}{2} \left[1 - \frac{1}{\mathrm{e}} - \frac{1}{\mathrm{e}} - 2 \sum \limits_{n=0}^\infty \int \limits_{2 \pi n}^{2 \pi (n+1)} \left(\mathrm{e}^{-\mathrm{e}^{-t}-t} - \mathrm{e}^{-\mathrm{e}^{t}+t} \right) \left(\frac{t}{2\pi} - n\right) \, \mathrm{d} t \right] \\ &= \frac{\pi}{2} \left[1 - \frac{2}{\mathrm{e}} - \frac{1}{\pi} \int \limits_0^\infty \left[-\ln(x) \mathrm{e}^{-x}\right] \, \mathrm{d} x + 2 \sum \limits_{n=0}^\infty n \left(\mathrm{e}^{-\mathrm{e}^{-2\pi(n+1)}}-\mathrm{e}^{-\mathrm{e}^{-2\pi n}} - \mathrm{e}^{-\mathrm{e}^{2\pi n}} + \mathrm{e}^{-\mathrm{e}^{2\pi(n+1)}}\right) \right] \\ &\equiv \frac{\pi}{2}-\frac{\pi}{\mathrm{e}}- \frac{\gamma}{2} + \pi (S_1 - S_2) \, . \end{align} The second sum is essentially equal to zero: $$ S_2 = \sum \limits_{n=1}^\infty n \left( \mathrm{e}^{-\exp[2\pi n]} - \mathrm{e}^{-\exp[2\pi(n+1)]}\right) = \sum \limits_{n=1}^\infty \mathrm{e}^{- \exp(2 \pi n)} \simeq 3 \cdot 10^{-233} \, . $$ The first sum can be simplified using the exponential series: $$ S_1 = \sum \limits_{n=1}^\infty n \left( \mathrm{e}^{-\exp[-2\pi (n+1)]} - \mathrm{e}^{-\exp[-2\pi n]}\right) = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k!} \sum \limits_{n=1}^\infty \mathrm{e}^{-2 \pi k n} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k! (\mathrm{e}^{2 \pi k} - 1)} \, .$$ Keeping only the first three terms of $S_1$ and ignoring $S_2$ , we obtain $$ S \simeq \pi\left[\frac{1}{2} - \frac{1}{\mathrm{e}} - \frac{\gamma}{2 \pi} + \frac{1}{\mathrm{e}^{2\pi} - 1} - \frac{1}{2(\mathrm{e}^{4\pi} - 1)} + \frac{1}{6(\mathrm{e}^{6\pi} - 1)}\right] \simeq 0.13233339071\color{red}{3} \, .$$ The red three is the first deviation from the 'exact' value. I do not believe that there are closed-form expressions for $S_1$ and $S_2$ , but at least this result is not too far off.

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  • $\begingroup$ Impressive and beautiful solution ! $\endgroup$ Feb 4, 2019 at 7:15
  • $\begingroup$ @ClaudeLeibovici Thank you! $\endgroup$ Feb 5, 2019 at 20:11
  • $\begingroup$ It's very clever to use the fourier series, and this is a beautiful representation ! $\endgroup$ Feb 5, 2019 at 20:51
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May be you can evaluate $\int_{0}^{\infty}{e^{-x+in\ln{x}}dx}=\int_{0}^{\infty}{x^{in}e^{-x}dx}$ and take imaginary part. It looks like Euler-$\Gamma$ something...

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  • $\begingroup$ Actually, that's where I'm coming from, but I don't think there is a known closed-form expression for $\Gamma(1+in)$... $\endgroup$ Feb 4, 2019 at 0:17

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