# Evaluating $\int_{-\infty}^{\infty}\frac{e^{-\alpha x}}{(1+e^{-x})^{1+\lambda}}\,\mathrm{d}x$ for $\alpha,\lambda>0$

For $$\alpha>0,\lambda>0$$, I am trying to evaluate the integral

$$I(\alpha,\lambda)=\int_{\mathbb R}\frac{e^{-\alpha y}}{(1+e^{-y})^{1+\lambda}}\,\mathrm{d}y$$

Basically I am struggling to find a proper substitution here.

I can write

$$I(\alpha,\lambda)=\int_{\mathbb R}e^{(1-\alpha)y}\frac{e^{-y}}{(1+e^{-y})^{1+\lambda}}\,\mathrm{d}y$$

Changing variables $$z=\frac{1}{(1+e^{-y})^{(1+\lambda)/2}}$$

, so that $$\mathrm{d}z=\left(\frac{1+\lambda}{2}\right)(1+e^{-y})^{\frac{\lambda-1}{\lambda+1}}\frac{e^{-y}}{(1+e^{-y})^{1+\lambda}}\,\mathrm{d}y\tag{1}$$

Now,

\begin{align} &\qquad\quad z(1+e^{-y})^{(1+\lambda)/2}=1 \\&\implies \ln z+\left(\frac{1+\lambda}{2}\right)\ln(1+e^{-y})=0 \\&\implies \ln(1+e^{-y})=-\frac{2}{1+\lambda}\ln z \\&\implies e^{-y}=\exp\left(-\frac{2}{1+\lambda}\ln z\right)-1=z^{-2/(1+\lambda)}-1 \end{align}

From $$(1)$$, I get $$\frac{e^{-y}}{(1+e^{-y})^{1+\lambda}}\,\mathrm{d}y=\left(\frac{2}{1+\lambda}\right)z^{\frac{\lambda-1}{\lambda+1}}\,\mathrm{d}z$$

And

$$e^{(1-\alpha)y}=(e^{-y})^{-(1-\alpha)}=\left(z^{-2/(1+\lambda)}-1\right)^{\alpha-1}$$

So finally,

$$I(\alpha,\lambda)=\frac{2}{1+\lambda}\int_0^1 \left(z^{-2/(1+\lambda)}-1\right)^{\alpha-1}z^{\frac{\lambda-1}{\lambda+1}}\,\mathrm{d}z$$

I transformed the domain of integration to $$(0,1)$$ so that I can use the Beta function and hence get an answer in terms of Gamma functions. But looks like I have made an error somewhere.

I also tried writing $$\frac{1}{(1+e^{-y})^{1+\lambda}}=\sum_{j=0}^\infty \binom{-(1+\lambda)}{j}e^{-yj}$$

, so that

\begin{align} I(\alpha,\lambda)&=\int_{\mathbb R} e^{-\alpha y}\sum_{j=0}^\infty \binom{-(1+\lambda)}{j}e^{-yj}\,\mathrm{d}y \\&\stackrel{?}{=}\sum_{j=0}^\infty \binom{-(1+\lambda)}{j}\int_{\mathbb R}e^{-(\alpha+j)y}\,\mathrm{d}y \end{align}

But this is not correct either.

According to Mathematica, I should get

$$I(\alpha,\lambda)=\frac{\Gamma(\alpha)\Gamma(1+\lambda-\alpha)}{\Gamma(1+\lambda)}\quad,\text{ if }\alpha<1+\lambda$$

You can use $$x = \frac{1}{1+\mathrm{e}^{-y}} ~ \Leftrightarrow ~ \mathrm{e}^{-y} = \frac{1-x}{x}$$ here. Then $$I(\alpha, \lambda) = \int \limits_0^1 x^{\lambda-\alpha} (1-x)^{\alpha-1} \, \mathrm{d} x = \operatorname{B} (1+\lambda-\alpha,\alpha) = \frac{\Gamma(\alpha) \Gamma(1+\lambda-\alpha)}{\Gamma(1+\lambda)} \, .$$