Finding the value of $\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$

Question

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.

Answer 1

An Upper Bound

First, note that $$ \begin{align} \lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x &\le\lim_{a\to\infty}\int_0^1ax^a\,\mathrm{d}x\tag{1a}\\ &=\lim_{a\to\infty}\frac{a}{a+1}\tag{1b}\\[3pt] &=1\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: $a^x\le a$
$\text{(1b)}$: evaluate the integral
$\text{(1c)}$: evaluate the limit

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One Approach to the Lower Bound

For any $b\in(0,1]$: $$ \begin{align} \lim_{a\to\infty}\int_{1-b}^1a^xx^a\,\mathrm{d}x &=\lim_{a\to\infty}\int_0^ba^{1-x}(1-x)^a\,\mathrm{d}x\tag{2a}\\ &\ge\lim_{a\to\infty}a\int_0^be^{-x\left(\frac{a}{1-b}+\log(a)\right)}\,\mathrm{d}x\tag{2b}\\ &=\lim_{a\to\infty}\frac{a}{\frac{a}{1-b}+\log(a)}\left(1-e^{-b\left(\frac{a}{1-b}+\log(a)\right)}\right)\tag{2c}\\[3pt] &=1-b\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: substitute $x\mapsto1-x$
$\text{(2b)}$: $-\frac{x}{1-x}\le\log(1-x)\le-x$; thus, on $[0,b]$,
$\phantom{\text{(3a):}}$ $-\frac{x}{1-b}\le\log(1-x)\le -x$
$\text{(2c)}$: integrate
$\text{(2d)}$: evaluate the limit

Thus, for any $b\in(0,1]$: $$ 1-b\le\lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x\le1\tag3 $$ Therefore, $$ \lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x=1\tag4 $$

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A Different Approach to the Lower Bound

Let $\epsilon\gt0$. Note that for $\lambda=1+\frac{\log(\epsilon)}{a+1}\le\epsilon^{\frac1{a+1}}$ $$ \begin{align} \int_\lambda^1x^a\,\mathrm{d}x &=\frac{1-\lambda^{a+1}}{a+1}\tag{5a}\\ &\ge\frac{1-\epsilon}{a+1}\tag{5b} \end{align} $$ Explanation:
$\text{(5a)}$: integrate
$\text{(5b)}$: $\lambda\le\epsilon^{\frac1{a+1}}$

For $x\in[\lambda,1]$, $$ \begin{align} a^x &\ge a^\lambda\tag{6a}\\ &=ae^{\frac{\log(\epsilon)\log(a)}{a+1}}\tag{6b}\\ &\ge a\left(1+\frac{\log(\epsilon)\log(a)}{a+1}\right)\tag{6c} \end{align} $$ Explanation:
$\text{(6a)}$: $x\ge\lambda$
$\text{(6b)}$: $a^\lambda=e^{\log(a)\left(1+\frac{\log(\epsilon)}{a+1}\right)}$
$\text{(6c)}$: $e^x\ge1+x$

Thus, $$ \begin{align} \lim_{a\to\infty}\int_\lambda^1a^xx^a\,\mathrm{d}x &\ge\lim_{a\to\infty}a\left(1+\frac{\log(\epsilon)\log(a)}{a+1}\right)\frac{1-\epsilon}{a+1}\tag{7a}\\ &=1-\epsilon\tag{7b} \end{align} $$ Explanation:
$\text{(7a)}$: apply the lower bounds from $(5)$ and $(6)$
$\text{(7b)}$: take the limit

Therefore, since $(1)$ provides an upper bound of $1$ and $(7)$ provides a lower bound of $1-\epsilon$ for any $\epsilon\gt0$, we have $$ \lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x=1\tag8 $$

Answer 2

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 $a>e$ $$\begin{align} I_a&=\int^1_0a^{1-u}(1-u)^a\,du\geq a\int^1_0a^{-u}\exp\big(\tfrac{-au}{1-u}\big)\,du\\ &\geq a\int^{1/\log a}_0e^{-u\big(\log a+\tfrac{a}{1-u}\big)}\,du\\ &\geq a\int^{1/\log a}_0e^{-u\big(\log a+\tfrac{a}{1-1/\log a}\big)}\,du\\ &=\frac{1}{\frac{\log a}{a}+\frac{1}{1-1/\log a}}\Big(1-e^{-\tfrac{1}{\log a}\big(\log a+\tfrac{a}{1-1/\log a}\big)}\Big)\xrightarrow{a\rightarrow\infty}1 \end{align} $$

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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: $$I_a\geq \int^{1/a}_0a^{1-u}(1-au)\,du\xrightarrow{a\rightarrow\infty}\frac12$$