Study the convergence of the improper integral without calculating it, then find its value when it converge

Question

Integral : $\large\int_{-\infty}^{+\infty}\frac{1}{e^x+e^{-x}+1}dx$

I am teaching myself calculus III , and having a problem solving such exercises.How can I study the convergence of integral that has exponential in it

Answer 1

The hyperbolic cosine is an even function, hence: $$\int_{-\infty}^{+\infty}\frac{dx}{e^x+e^{-x}+1}=2\int_{0}^{+\infty}\frac{dx}{e^x+e^{-x}+1}.$$ If now we set $x=\log t$, we end with: $$\int_{-\infty}^{+\infty}\frac{dx}{e^x+e^{-x}+1}=2\int_{1}^{+\infty}\frac{dt}{t^2+t+1}=2\int_{\frac{3}{2}}^{+\infty}\frac{dt}{t^2+3/4},$$ and by putting $t=\frac{\sqrt{3}}{2}z$, we have: $$\int_{-\infty}^{+\infty}\frac{dx}{e^x+e^{-x}+1}=\frac{4\sqrt{3}}{3}\int_{\sqrt{3}}^{+\infty}\frac{dz}{z^2+1}=\frac{4\sqrt{3}}{3}\arctan{\frac{1}{\sqrt{3}}}=\frac{2\pi\sqrt{3}}{9}.$$

For the convergence, it is sufficient to notice that: $$0< \frac{1}{1+e^{x}+e^{-x}}\leq \frac{1}{e^{|x|}},$$ hence the integral is positive and bounded by: $$\int_{-\infty}^{+\infty}e^{-|x|}dx = 2.$$