# A limit about $\int_0^\infty x^{-x}e^{tx}dx$

How to prove $$\lim_{t\to +\infty}\frac{\int_0^{+\infty} x^{-x}e^{tx}dx}{e^{\frac12(t-1)+e^{t-1}}}=\sqrt{2\pi}.$$ Someone asked this difficult question, I have tried Taylor formula of $e^x$ but failed, could you show me the method?

• Hint: What famous mathematical formula contains in itself all of the following three expressions: $x^x,$ $e^x,$ and $\sqrt{2\pi}$ ? – Lucian Dec 15 '15 at 12:25
• @Lucian Gamma function, stirling formula? – Eufisky Dec 15 '15 at 14:41

## 1 Answer

Writing $x^x = \exp(x \ln x)$, your integral is $$\int_0^\infty \exp((t - \ln x)x) \; dx$$ The maximum of $(t - \ln x) x$ with respect to $x > 0$ occurs at $x_0 = \exp(t-1)$, and taking a Taylor series around that point

$$(t - \ln x)(x) = x_0 - \dfrac{(x - x_0)^2}{2 x_0} + O((x-x_0)^3)$$

Thus (after taking care of some details) your integral is asymptotic as $t \to \infty$ to $$\int_{-\infty}^\infty \exp\left(x_0 - \dfrac{(x - x_0)^2}{2 x_0}\right)\; dx = \sqrt{2\pi} \exp\left(e^{t-1} + t/2 - 1/2\right)$$