How do I evaluate $\int\limits_{0}^{+\infty} \frac{e^{-\frac{1}{x}}}{x^\alpha(1+x^\alpha)} \mathrm dx$, where $\alpha \in \mathbb{R}$

Question

\begin{align*} \int_{0}^{+\infty} \frac{e^{-\frac{1}{x}}}{x^\alpha(1+x^\alpha)} \mathrm dx &= \int_{0}^{+\infty} e^{-\frac{1}{x}}\left(\frac{1+x^\alpha}{x^\alpha(1+x^\alpha)} - \frac{x^\alpha}{x^\alpha(1+x^\alpha)}\right)dx\\ &= \int_{0}^{+\infty} \frac{e^{-\frac{1}{x}}}{x^\alpha} \mathrm dx - \int_{0}^{+\infty} \frac{e^{-\frac{1}{x}}}{1+x^\alpha} \mathrm dx \end{align*}

So I get :

$$\int_{0}^{+\infty} \frac{e^{-\frac{1}{x}}}{x^\alpha} \mathrm dx = \Gamma(\alpha-1)$$

But I don´t know how to evaluate :

$$- \int_{0}^{+\infty} \frac{e^{-\frac{1}{x}}}{1+x^\alpha} \mathrm dx$$

Thanks in advance for your replies.

Answer 1

With the substitution of $x\rightarrow \frac{1}{y}$ the integral

$$\int_{0}^{\infty }\exp \left( -y\right) \frac{y^{\alpha -2}}{1+y^{\alpha }}dy$$

can be interpreted as a laplace transformation of

$$\mathcal{L}\left\{ \frac{y^{\alpha -2}}{1+y^{\alpha }}\right\} =\mathcal{L}% \left\{ y^{\alpha -2}G_{1,1}^{1,1}\left( y^{\alpha }\left\vert \begin{array}{c} 0 \\ 0% \end{array}% \right. \right) \right\} $$ for $s=1$. Here the integrand is expressed in form of a Meijer $G$ - function. With the help of Mathai (2.29,p. 52) the solution can be expressed

$$\mathcal{L}\left[ y^{\alpha -2}G_{1,1}^{1,1}\left( y^{\alpha }\left\vert \begin{array}{c} 0 \\ 0% \end{array}% \right. \right) \right] \overset{s\rightarrow 1}{=}H_{3,2}^{1,3}\left( 1\left\vert \begin{array}{c} \left( 0,1\right) ,\left( 0,1\right) ,\left( 2-\alpha ,\alpha \right) \\ \left( 0,1\right) ,\left( 0,1\right) \end{array}% \right. \right) $$ by a $H$ - Fox function.