TITLE: Finding the value of $\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$
QUESTION [7 upvotes]: I'm trying to find the value of $$\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$$
My attempt:
Let $\epsilon  >0$ be given.
$ x\mapsto a^{x}$ is continuous at $ 1$ so there is a $d_a\in ( 0,1)$ such that $|a^{x} -a|< \epsilon $ for all $ x\in [d_a,1]$. WLOG, let $d_a<1/2$.
$ |\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} |=|\int _{0}^{1}\left( a^{x} -a\right) x^{a} \ dx|\leq |\int _{0}^{d}\left( a^{x} -a\right) x^{a} \ dx|+|\int _{d}^{1}\left( a^{x} -a\right) x^{a} \ dx|$
\begin{align*}
\left|\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} \right| & \leq \left|\int _{0}^{d_a}\left( a^{x} -a\right) x^{a} \ dx\right|+\left|\int _{d_a}^{1}\left( a^{x} -a\right) x^{a} \ dx\right|\\
 & \leq \int _{0}^{d_a}\left( a -a^{x}\right) x^{a} \ dx+\epsilon \left|\int _{d_a}^{1} x^{a} \ dx\right|\\
 & \leq \int _{0}^{d_a}\left( a -a^{x}\right) x^{a} \ dx+\epsilon \\
 & \leq \int _{0}^{1/2} a(1/2)^{a} \ dx-a\int _{0}^{d_a} x^{a} dx+\epsilon \\
 & \leq a(1/2)^{a}  +\epsilon 
\end{align*}
$0\leq \lim _{a\rightarrow \infty }\inf |\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} |\leq \lim _{a\rightarrow \infty }\sup |\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} |\leq \epsilon $
Since this is true for every $\epsilon  >0,$it follows that $ \lim _{a\rightarrow \infty }\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} =0$.
Is my proof correct? Thanks.

REPLY [3 votes]: Here is a slightly different solution base on the change of variable $u=1-x$ that robjohn used in one of his solutions.
First the easy bound
$$I_a=\int^1_0e^{x\log a}x^a\,dx\leq a\int^1_0x^a\,dx=\frac{a}{1+a}\xrightarrow{a\rightarrow\infty}1$$
The change of variables $u=1-x$ yields
$$I_a=\int^1_0a^{1-u}(1-u)^a\,du$$
Since $e^{\tfrac{u}{1-u}}\geq 1+\frac{u}{1-u}=\frac{1}{1-u}$, we have that $(1-u)^a\geq e^{-\tfrac{au}{1-u}}$ for $0\leq u \leq 1$ (both RHS and LHS take value $0$ when $u\rightarrow1-$). Hence, for all $a>e$
$$\begin{align}
I_a&=\int^1_0a^{1-u}(1-u)^a\,du\geq  a\int^1_0a^{-u}\exp\big(\tfrac{-au}{1-u}\big)\,du\\
&\geq  a\int^{1/\log a}_0e^{-u\big(\log a+\tfrac{a}{1-u}\big)}\,du\\
&\geq a\int^{1/\log a}_0e^{-u\big(\log a+\tfrac{a}{1-1/\log a}\big)}\,du\\
&=\frac{1}{\frac{\log a}{a}+\frac{1}{1-1/\log a}}\Big(1-e^{-\tfrac{1}{\log a}\big(\log a+\tfrac{a}{1-1/\log a}\big)}\Big)\xrightarrow{a\rightarrow\infty}1
\end{align}
$$

Comment: the bound $(1-u)^a\geq e^{-\tfrac{au}{1-u}}$ seems to be optimal to get the right limit. Other bounds, for example by using Bernoulli's inequality $(1-u)^a\geq 1-au\geq0$ for $0\leq u\leq 1/a$, fall short:
$$I_a\geq \int^{1/a}_0a^{1-u}(1-au)\,du\xrightarrow{a\rightarrow\infty}\frac12$$