Is it possible to calculate the following definite integral in a closed form?
$$ \int_0^\infty \left| \sin x \cdot \sin (\pi x) \right| e^{-x} \, dx$$
[Math] Closed form for improper definite integral involving trig functions and exponentials
closed-formdefinite integrals
Related Solutions
Let us denote the integral by $I_n$: $$ I_n \equiv \int_0^1\,_2F_1\left(-n,1+n;1;x^3\right) \left[3 \, _2F_1\left(1-n,2+n;1;x^3\right)-\, _2F_1\left(1-n,2+n;2;x^3\right)\right]\, x \, \mathrm{d}x\,. $$ We first expand both hypergeometric functions on the right of the integral using the series representation around $x=0$. This yields
$$ I_n=\int_0^1 \sum_{j=0}^{+\infty}\frac{3 j x^{3 j+1} \Gamma (j+1) \, _2F_1\left(-n,n+1;1;x^3\right) (1-n)_j (n+2)_j}{j! (1)_j (2)_j} \mathrm{d}x=\sum_{j=0}^{+\infty}\int_0^1 \frac{3 j x^{3 j+1} \Gamma (j+1) \, _2F_1\left(-n,n+1;1;x^3\right) (1-n)_j (n+2)_j}{j! (1)_j (2)_j} \mathrm{d}x $$
The resulting integral can then be expressed in terms of $_3 F_2$ generalised hypergeometric functions:
$$ I_n = \sum_{j=0}^{+\infty} \frac{3 j \Gamma (j+1) (1-n)_j (n+2)_j \, _3F_2\left(j+\frac{2}{3},-n,n+1;1,j+\frac{5}{3};1\right)}{(3 j+2) j! (1)_j (2)_j}\,. $$
Now one can use the Pfaff–Saalschütz Balanced Sum (see for instance https://dlmf.nist.gov/16.4) to express the $_3F_2$ at unit value in terms of more Pochhammer symbols. In particular, we get
$$ I_n = -\sum_{j=0}^{+\infty} \frac{\Gamma \left(j+\frac{2}{3}\right) (1-n)_j \Gamma \left(-j+n+\frac{1}{3}\right) \Gamma (j+n+2)}{\Gamma \left(-j-\frac{2}{3}\right) \Gamma (j+1) \Gamma (j+2) \Gamma (n+2) \Gamma \left(j+n+\frac{5}{3}\right)} $$
which one recognises as being related to a $_4 F_3$ generalised hypergeometric function evaluated at unit argument. After some simplifications, one finds
$$ I_n=\frac{\Gamma \left(\frac{5}{3}\right) \Gamma \left(n+\frac{1}{3}\right)}{\Gamma \left(\frac{1}{3}\right) \Gamma \left(n+\frac{5}{3}\right)} \, _4F_3\left(\frac{2}{3},\frac{5}{3},1-n,n+2;2,\frac{2}{3}-n,n+\frac{5}{3};1\right)\,. $$
With CAS help and Mellin Transform:
$$\int_0^1 \frac{\exp \left(\frac{1}{\log (x)}\right)}{\sqrt[5]{x} \sqrt[5]{\log (x)}} \, dx=\\\mathcal{M}_a^{-1}\left[\int_0^1 \mathcal{M}_a\left[\frac{\exp \left(\frac{a}{\log (x)}\right)}{\sqrt[5]{x} \sqrt[5]{\log (x)}}\right](s) \, dx\right](1)=\\\mathcal{M}_a^{-1}\left[\int_0^1 \frac{(-1)^{-s} \Gamma (s) \log ^{-\frac{1}{5}+s}(x)}{\sqrt[5]{x}} \, dx\right](1)=\\\mathcal{M}_s^{-1}\left[-(-1)^{4/5} \left(\frac{4}{5}\right)^{-\frac{4}{5}-s} \Gamma (s) \Gamma \left(\frac{4}{5}+s\right)\right](1)=\\-(-1)^{4/5} *\sqrt[5]{2}* 5^{2/5} *K_{\frac{4}{5}}\left(\frac{4}{\sqrt{5}}\right)\approx0.302595\, -0.219848 i$$
Where:
$\mathcal{M}_a[f(a)](s)$ is Mellin Transform,
$\mathcal{M}_s^{-1}[f(s)](a)$ is Inverse Mellin Transform,
$K_{\frac{4}{5}}\left(\frac{4}{\sqrt{5}}\right)$ is modified Bessel function of the second kind.
Best Answer
One could give the following a try: Develop $|\sin x|$ into a Fourier series. You get $$|\sin x|={2\over\pi}-{4\over\pi}\sum_{k=1}^\infty {1\over 4k^2 -1}\cos(2kx)\ .$$ Similarly $$|\sin (\pi x)|={2\over\pi}-{4\over\pi}\sum_{k=1}^\infty {1\over 4k^2 -1}\cos(2\pi kx)\ .$$ Since the two series are absolutely convergent you can multiply them, obtaining a double series of the form $$\sum_{k,l} 2c_{k,l}\cos(2kx)\cos(2\pi l x)=\sum_{k,l} c_{k,l}\bigl(\cos \bigl((2(k+\pi l)x\bigr)+\cos\bigl(2(k-\pi l) x\bigr)\bigr)\ .$$ Now $$\int_0^\infty \cos(q x)e^{-x}\ dx={1\over 1+q^2}\ ;$$ therefore you will end up with a huge double series containing terms of the form $${c\over (4k^2-1)(4l^2-1)\bigl(1+4(k\pm \pi l)^2\bigr)}\ .$$ I wish you luck$\ldots$