How do you find the integral of #int dx/(e^x +e^-x)# from negative infinity to infinity?

Question

How do you find the integral of #int  dx/(e^x +e^-x)# from negative infinity to infinity?

Christopher Agresta · Accepted Answer

#pi/2# #f(x)=1/(e^x+e^(-x))# #f(-x)=1/(e^(-x)+e^(x))# =>f(x)=f(-x) => the given function is even function son#int_-oo^oo1/(e^x+e^(-x))dx=2int_0^oo1/(e^x+e^(-x))dx=2int_0^ooe^x/(e^(2x)+1)dx=2int_0^ooe^x/((e^x)^2+1)dx# if #y =tan^-1x# #dy=1/(x^2+1)dx# =>#y=int1/(x^2+1)dx=tan^-1x# if you substitute #x=e^x# then #dtan^-1(e^x)=1/((e^x)^2+1)e^xdx# =>#2int_0^ooe^x/((e^x)^2+1)dx=2int_0^ood(tan^-1e^(x))dx=2[tan^-1e^x]_0^oo=2[tan^-1oo-tan^-1(1)]=2[pi/2-pi/4]=pi/2#

Axel Alvarez · Answer

undefined To find the integral of $ \int \frac{{dx}}{{e^x + e^{-x}}} $ from negative infinity to infinity, we can use a method called contour integration. We'll integrate along a contour in the complex plane and then let the radius of the contour tend to infinity. This integral can be evaluated by using residues.

The function $ \frac{{1}}{{e^x + e^{-x}}} $ can be rewritten in terms of hyperbolic functions as $ \frac{{\cosh(x)}}{{\sinh(x) \cosh(x)}} = \frac{{1}}{{\sinh(x)}} $.

By applying contour integration techniques, we integrate $ \frac{{1}}{{\sinh(x)}} $ over a semicircular contour in the upper half-plane, and then let the radius of the contour tend to infinity. The integral along the straight line segment along the real axis from $ -R $ to $ R $ will be the integral we're interested in. As the radius tends to infinity, the contribution from the circular part of the contour will tend to zero.

The residue of $ \frac{{1}}{{\sinh(x)}} $ at its pole $ x = 0 $ is $ 1 $.

By applying the residue theorem, the integral along the semicircular contour is $ 2\pi i \times 1 = 2\pi i $.

Therefore, the integral of $ \frac{{dx}}{{e^x + e^{-x}}} $ from negative infinity to infinity is $ 2\pi i $.