# Need help to evaluate $I_n=\int_{0}^{\infty}e^{-x}\sin(n\ln(x))dx$

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 ?

• Mathematica gives the following result: $I_n = \frac{1}{2} i (\Gamma (1-i n)-\Gamma (i n+1))$.
– JimB
Feb 4, 2019 at 0:14
• 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.)
– JimB
Feb 4, 2019 at 4:17

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 $$\sum_{n=1}^\infty \frac{I_n}{n}=\sum_{n=1}^\infty \frac{\Gamma(in)+\Gamma(-in)}{2}=\sum_{n=1}^\infty \frac{\Gamma(in)+\overline{\Gamma(in)}}{2}=\sum_{n=1}^\infty \Re{(\Gamma(in))}$$

• Thank you very much, this is a nice representation ! Feb 5, 2019 at 20:51

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.

• Impressive and beautiful solution ! Feb 4, 2019 at 7:15
• @ClaudeLeibovici Thank you! Feb 5, 2019 at 20:11
• It's very clever to use the fourier series, and this is a beautiful representation ! Feb 5, 2019 at 20:51

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...

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