I am looking for Analytic solution to a definite integral. Or an approriate transformation to apply. the conditions on $\alpha$ , $\beta$ being positive real numbers while $n$ is positive integer.the integral is given as $$ \int_{-\infty}^{\infty}x^ne^{-\beta x}\left(1 + \alpha e^{-\beta x}\right)^{-1/\alpha}\,dx $$

  • $\begingroup$ When I see integrals of a form simillar to $\int_{-\infty}^{+\infty}e^{-x}dx$ I always have to think about the residual theorem. but then there are some conditions on $\alpha$ and $\beta$ nescessairy to make sure the function gets fast enough sufficiently small. ("connect" (approximately) $\infty$ and $-\infty$ such that your integral is defined on a closed curve in $\mathbb{C}$) $\endgroup$
    – Max
    Mar 19, 2014 at 9:15
  • $\begingroup$ @Max Conditions on just $\alpha$ actually. The dependency on $\beta$ can be pulled outside the integral via a simple change of scale. $\endgroup$
    – David H
    Mar 24, 2014 at 11:20

1 Answer 1


Consider $\int_{-\infty}^\infty e^{-\beta x}\left(1+\alpha e^{-bx}\right)^{-\frac{1}{\alpha}}~dx$ ,

$\int_{-\infty}^\infty e^{-\beta x}\left(1+\alpha e^{-bx}\right)^{-\frac{1}{\alpha}}~dx$

$=\int_\infty^0x^\frac{\beta}{b}~(1+\alpha x)^{-\frac{1}{\alpha}}~d\left(-\dfrac{\ln x}{b}\right)$

$=\dfrac{1}{b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+\alpha x)^{-\frac{1}{\alpha}}~dx$


$=\dfrac{1}{\alpha^\frac{\beta}{b}~b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+x)^{-\frac{1}{\alpha}}~dx$


$\therefore\int_{-\infty}^\infty x^ne^{-\beta x}\left(1+\alpha e^{-\beta x}\right)^{-\frac{1}{\alpha}}~dx=(-1)^n\dfrac{d^n}{d\beta^n}\left(\dfrac{1}{\alpha^\frac{\beta}{b}~b}B\left(\dfrac{\beta}{b},\dfrac{1}{\alpha}-\dfrac{\beta}{b}\right)\right)(b=\beta)$

  • $\begingroup$ I think there is some misunderstanding here. Its exp^(-beta*x) inside the bracket as well as outside it. $\endgroup$
    – SA-255525
    Apr 17, 2014 at 17:37

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