It seems that there are real integrals that are immune to all real methods of integration and one has to apply the residue theorem and contour integration. Here is my collection
$$\int_0^1 x^{-x}(1-x)^{x-1}\sin \pi x\,\mathrm{d}x=\frac{\pi}{e}\tag {1}$$
$$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+W(1)}=\frac{1}{1+\Omega}\tag {2}$$
$$\int^{\pi/2}_{0}\cos(xt)\cos^y(t)\,\mathrm{d}t=\frac{\pi \Gamma(x+1)}{2^{y+1}\Gamma\left(\frac{x+y+2}{2}\right)\Gamma\left(\frac{2-x+y}{2}\right)}\tag{3}$$
$$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{\mathrm{d}x}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8} \tag {4}$$
$$\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{(e^x+x+1)^2+\pi^2}=\frac{2}{3}\tag {5}$$
Are there any other examples ?
References
- https://artofproblemsolving.com/community/c7h501365p2817263
- Interesting integral related to the Omega Constant/Lambert W Function
- http://advancedintegrals.com/2017/04/integrating-a-function-around-three-branches-using-a-semi-circle-contour/
- https://arxiv.org/pdf/1402.3830.pdf
- Integral: $\int_{-\infty}^{\infty} \frac{dx}{(e^x+x+1)^2+\pi^2}$
