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.
Best Answer
An Upper Bound
For $a\ge1$, $$ \begin{align} \int_0^1a^xx^a\,\mathrm{d}x &\le\int_0^1ax^a\,\mathrm{d}x\tag{1a}\\ &=\frac{a}{a+1}\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $a^x\le a$
$\text{(1b)}$: evaluate the integral
A Lower Bound $$ \begin{align} \int_0^1a^xx^a\,\mathrm{d}x &=\frac{a}{a+1}\int_0^1a^{x-1}\,(a+1)x^a\,\mathrm{d}x\tag{2a}\\ &\ge\frac{a}{a+1}a^{\int_0^1(x-1)\,(a+1)x^a\,\mathrm{d}x}\tag{2b}\\ &=\frac{a}{a+1}a^{-\frac1{a+2}}\tag{2c} \end{align} $$ Explanation:
$\text{(2a):}$ factor $\frac{a}{a+1}$ out of the integral
$\text{(2b):}$ Jensen's Inequality
$\phantom{\text{(2b):}}$ $a^x$ is convex and $\int_0^1(a+1)x^a\,\mathrm{d}x=1$
$\text{(2c):}$ evaluate the integral
Squeeze The Limit
For $a\ge1$, $(1)$ and $(2)$ give $$ \frac{a}{a+1}a^{-\frac1{a+2}}\le\int_0^1a^xx^a\,\mathrm{d}x\le\frac{a}{a+1}\tag3 $$ and the Squeeze Theorem gives $$ \lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x=1\tag4 $$ Here is a graph plotting the bounds given in $(3)$ and also the lower bound, in red, given in Paramanand Singh's nice answer (both our upper bounds are the same):