# Integrals that can't be solved by real methods

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. – Zaid Alyafeai Apr 24 '17 at 7:29
• In the third integral, if m,n>0 are integers, no need contour integration imho – FDP Apr 24 '17 at 8:34
• math.stackexchange.com/questions/1055468/… – Ron Gordon Apr 24 '17 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 – tired Apr 24 '17 at 17:32
• @TZakrevskiy See This Answer in which I presented a real analysis method for evaluating the Fourier Cosine Transform $\int_{-\infty}^\infty \frac{\cos(\omega x)}{1+x^2}\,dx$. – Mark Viola Apr 24 '17 at 17:41