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\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.

Gaetano D. Agostino
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  • Related: https://math.stackexchange.com/questions/3042335/seeking-methods-to-solve-int-0-infty-frace-xnxn-1-dx?noredirect=1&lq=1 – mrtaurho Feb 13 '19 at 16:54
  • It does look good regarding a closed-form. Experimenting, especially after enforcing the substitution $1/x\mapsto x$, seems that for $\alpha=2$ there is some kind of closed-form involving the Sine and the Cosine Integral. However, for $\alpha=3$ and greater WolframAlpha does not produce anything at all. I do not even know how to get started for $\alpha\notin\mathbb N$. – mrtaurho Feb 13 '19 at 17:00
  • If that helps, the integral can be evaluated in terms of the Fox H-function in the same way as here. – Maxim Feb 13 '19 at 20:30
  • @Maxim - was just about to link my question! :-) –  Feb 14 '19 at 11:02
  • I found the solution, but have no time to test it. First substitution 1/x->x, then express $\frac{1}{x^{2}}\frac{x^{\alpha }}{x^{\alpha }+1}$ by a MeijerG-Function. Then set: $\alpha = l/k$ and use http://functions.wolfram.com/HypergeometricFunctions/MeijerG/21/02/07/0005/. If you need the general solution just change the MeijerG - Function by a H-Fox-Function https://www.researchgate.net/publication/266566090_The_H-function_Theory_and_Applications. If you need more help, please let me know – stocha Feb 14 '19 at 17:19

1 Answers1

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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.

stocha
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    The change of variables gives $e^{-y} y^{2 \alpha - 2}/(1 + y^\alpha)$. Also, notice that the H-function can be simplified to $H_{2, 1}^{1, 2}$ by cancelling two identical factors. – Maxim Feb 16 '19 at 12:40
  • @Maxim you are right, thank you very much for this note. First I consider the Plancherel theorem and the laplace of Dotsenko function and had not much time left. I also consider the last integral of the question after "how to evaluate:" – stocha Feb 16 '19 at 17:08
  • I see, I was thinking about the integral in the title. So the answer for that integral is $\Gamma(\alpha - 1)$ minus your final result, with the condition $\alpha > 1$. Then you can argue that this still holds for $\alpha > 1/2$ by analytic continuation. – Maxim Feb 16 '19 at 18:08