# Integrals that can't be solved by real methods [closed]

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

• @Masacroso, of course but you won't get that nice closed form solution like the first one using approximations. My question isn't about integrals that can't be expresses using elementary function's rather those where contour integrations seem superior. Apr 24, 2017 at 7:29
• In the third integral, if m,n>0 are integers, no need contour integration imho
– FDP
Apr 24, 2017 at 8:34
• math.stackexchange.com/questions/1055468/… Apr 24, 2017 at 11:52
• @ZaidAlyafeai i'm quite confident that $(3)$ can be done without contour integrals/residue theorem. Furthermore math.stackexchange.com/questions/253910/… might ne intersting for you Apr 24, 2017 at 17:32
• @TZakrevskiy See This Answer in which I presented a real analysis method for evaluating the Fourier Cosine Transform $\int_{-\infty}^\infty \frac{\cos(\omega x)}{1+x^2}\,dx$. Apr 24, 2017 at 17:41