Solution to the definite integral $\int_0^{\infty}x^{-\alpha} e^{-\beta/x}\,dx$

Question

Is there a explicit solution to this definite integral, with $\alpha>0$ and $\beta>0$ :

$$\int_0^{\infty}x^{-\alpha} e^{-\beta/x}\,dx$$

If yes, what is it?

Answer 1

Substitute $\beta/x=t$, so you get $$ \int_0^\infty (t/\beta)^{\alpha}e^{-t}\frac{\beta}{t^2}\,dt =\beta^{1-\alpha}\int_0^\infty t^{\alpha-2}e^{-t}\,dt=\beta^{1-\alpha}\Gamma(\alpha-1) $$ because $$ \Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt $$ See Gamma function on Wikipedia

Note that this is undefined for $0<\alpha\le1$, so your assumption should be $\alpha>1$.

Answer 2

The method prescribed by egreg is far superior to what I'm going to put forward. Here we will address your integral: \begin{equation} I\left(\alpha, \beta \right) = \int_0^\infty x^{-\alpha} e^{-\beta/x}\:dx\nonumber \end{equation} We now employ Fubini's Theorem and take the Laplace Transform with respect to $\beta$: \begin{align} \mathscr{L}_{\beta \rightarrow s}\left[ I\left(\alpha, \beta \right) \right] &= \int_0^\infty x^{-\alpha} \frac{1}{s - -\frac{1}{x}}\:dx = \int_0^\infty \frac{x^{1 - \alpha}}{xs + 1}\:dx = \frac{1}{s} \int_0^\infty \frac{x^{1 - \alpha}}{x + s^{-1}}\:dx \end{align} We now use the solution that I derived here: \begin{equation} \int_0^\infty \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) \end{equation} Thus, \begin{align} \mathscr{L}_{\beta \rightarrow s}\left[ I\left(\alpha, \beta \right) \right] &= \frac{1}{s} \int_0^\infty \frac{x^{1 - \alpha}}{x + s^{-1}}\:dx = \frac{1}{s} \cdot \frac{1}{1} \cdot \left(s^{-1}\right)^{\frac{1 - \alpha + 1 }{1} - 1}\cdot B\left(1 - \frac{1 - \alpha + 1}{1}, \frac{1 - \alpha + 1}{1} \right)\nonumber \\ &= s^{\alpha - 2} B\left(\alpha - 1, 2 - \alpha\right) \end{align} Where $B(\cdot, \cdot)$ is the Beta Function. We now employ the relationship between the Beta and Gamma function being: \begin{equation} B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} \end{equation} Thus, \begin{align} \mathscr{L}_{\beta \rightarrow s}\left[ I\left(\alpha, \beta \right) \right] &= s^{\alpha - 2} B\left(\alpha - 1, 2 - \alpha\right) = s^{\alpha - 2} \frac{\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha)}{\Gamma\left( \alpha - 1 + 2 - \alpha\right)} = s^{\alpha - 2}\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha) \end{align} We now take the inverse Laplace Transform with respect to $s$: \begin{align} I\left(\alpha, \beta \right)= \mathscr{L}_{s \rightarrow \beta}^{-1}\left[s^{\alpha - 2}\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha) \right] = \frac{\beta^{1 - \alpha}}{\Gamma(2 - \alpha) }\cdot\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha) =\beta^{1 - \alpha}\Gamma\left(\alpha - 1\right) \end{align} And thus, \begin{equation} I\left(\alpha, \beta \right) = \beta^{1 - \alpha}\Gamma\left(\alpha - 1\right) \nonumber \end{equation}