Use the $\Gamma$ function to solve $\int_{-\infty}^{\infty} e^{xt - e^{t}} dt$

Question

Use the $\Gamma$ function to solve $\int_{-\infty}^{\infty} e^{xt - e^{t}} dt$

Let $u=e^t$, then $du=e^t dt$, and $dt = \frac{du}{\ln|u|}$ because $t=\ln|u|$ so $xt = x\ln|u| = \ln|u^x|$, putting all this together our integral becomes:

$$\int_0^{\infty} e^{\ln|u^x|-u}\dfrac{du}{\ln|u|} = \int_0^\infty u^x e^{-u} \dfrac{du}{\ln|u|}$$

I'm not sure how to get rid of the $\ln|u|$ component to use the Gamma function, which is:

$$\Gamma(x):=\int_0^\infty e^{-t}t^{x-1}dt \quad \quad (x>0)$$

Also, I know that $\ln(1)=0$, so that I should probably split the integral at $1$ and use limits. Perhaps there is a less convoluted way to find this integral using the Gamma function. My professor also defined

$$\beta(x,y):=\int_0^1 t^{x-1}(1-t)^{y-1}dt \quad \quad (x,y > 0)$$

but I believe I only need the $\Gamma$-function for this problem.

Answer 1

Your second and third statements are contradictory. If $du=e^t dt$, then $dt=du/e^t=du/u$, not $du/\ln|u|$.