Krdytkyu

I need to solve the following integral.
$$int _{-infty }^{infty }::dfrac{e^{ax}}{e^x+1}dx$$

where $0<a<1$.

The substitution $e^x mapsto x$ gives

$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_0^infty frac{1}{x^{1-a}(1+x)}dx $$

where $0 < 1-a < 1$

We can use complex integration here. Due to the pole at $z=-1$, we pick a branch cut on the positive real axis such that $0 le arg(z) < 2pi$. The contour is a standard "keyhole contour" consisting of

• 
  $C_1$: left semicircle centered at $0$ with radius $epsilon$, going clockwise from $-epsilon i$ to $epsilon i$
  


• 
  $C_2$: straight line from $epsilon i$ to $R + epsilon i$
  


• 
  $C_3$: circle centered at $0$ with radius $R$, going counterclockwise from $R+iepsilon$ to $R - epsilon i$
  


• 
  $C_4$: straight line from $R - epsilon i$ to $-epsilon i$
  

In the limits $epsilon to 0$ and $R to infty$, the two straight line integrals converge to

$$ int_{C_2} f(z) dz to int_0^infty frac{1}{x^{1-a}e^{i0(1-a)}(1+x)}dx = I $$

$$ int_{C_4} f(z) dz to -int_0^infty frac{1}{x^{1-a}e^{i2pi(1-a)}(1+x)}dx = -e^{i2pi a}I $$

The remaining integrals should go to $0$

$$ leftvertint_{C_1} f(z)dzrightvert le frac{pi epsilon}{|z|^{1-a}|1+z|} le frac{pi epsilon^a}{1-epsilon} to 0 $$

$$ leftvertint_{C_3} f(z) dz rightvert le frac{2pi R}{|z|^{1-a}|1+z|} le frac{2pi R^a}{R-1} to 0 $$

Therefore

$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=-1} f(z) = -2pi i e^{ipi a} $$

So finally

$$ I = frac{2pi ie^{ipi a}}{e^{2pi a}-1} = frac{2pi i}{e^{ipi a}-e^{-ipi a}} = frac{pi}{sin(api)} $$

Edit: Here's a simpler method, credit to @Poujh

$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_{-infty}^infty f(x) dx $$

Take a rectangular contour:

• 
  $C_1$: Straight line from $-R$ to $R$
  


• 
  $C_2$: Straight line from $R$ to $R + i2pi$
  


• 
  $C_3$: Straight line from $R + i2pi$ to $-R + i2pi$
  


• 
  $C_4$: Straight line from $-R + i2pi$ to $-R$
  

In the limit of $Rtoinfty$, we have

$$ leftvertint_{C_2} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(R+iy)}}{e^{R+iy}+1} dyrightvert le 2pi leftvertfrac{e^{aR}}{e^R-1}rightvert to 0 $$

$$ leftvertint_{C_4} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(-R+iy)}}{e^{-R+iy}+1} rightvert le 2pi leftvert frac{e^{-aR}}{1-e^{-R}} rightvert to 0 $$

For the horizontal lines

$$ int_{C_1} frac{e^{az}}{e^z+1} dz to int_{-infty}^infty frac{e^{ax}}{e^x+1} dx = I $$

$$ int_{C_3} frac{e^{az}}{e^z+1} dz to -int_{-infty}^infty frac{e^{ax}e^{i2pi a}}{e^x+1} = -e^{i2pi a} I $$

We end up with the same result

$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=ipi} frac{e^{az}}{e^z+1} $$

where the residue is computed using the limit

$$ lim_{zto ipi} e^{az}frac{z-ipi}{e^z+1} = -e^{ipi a} $$