Generating Factorial Integral

Question

Prove that $\int_{-\infty}^{\infty}e^{at-e^t}\,dt=a!$ if $a \geq 1$.

How do I even go about proving this???

All I was able to do was confirm this for $a$ = $1,2,3...10$.

Answer 1

Your (improper) integral is not equal to $a!$.

By completing the square for the term in the exponent one can write $$at - t^2 = \frac{a^2}{4} - \left (t - \frac{a}{2} \right )^2.$$ So your integral becomes $$\int^\infty_{-\infty} e^{at - t^2} \, dt = e^{\frac{a^2}{4}} \int^\infty_{-\infty} e^{-(t - a/2)^2} \, dt.$$ After making the substitution $t \mapsto t - \frac{a}{2}$ one has $$\int^\infty_{-\infty} e^{at - t^2} \, dt = e^{\frac{a^2}{4}} \int^\infty_{-\infty} e^{-t^2} \, dt.$$ The result for this last integral is well known. It is the Gaussian (or probability) integral and is equal to $\sqrt{\pi}$. Thus $$\int^\infty_{-\infty} e^{at - t^2} \, dt = e^{\frac{a^2}{4}} \sqrt{\pi},$$ and is valid for all $a \in \mathbb{R}$.