Finding the value of $\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$ 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.
 A: Here is a slightly different solution base on the change of variable $u=1-x$.
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:
$$I_a\geq \int^{1/a}_0a^{1-u}(1-au)\,du\xrightarrow{a\rightarrow\infty}\frac12$$
A: Let's observe that $$\int_0^1(a-a^x)x^a\,dx$$ is slightly easier to handle. The integrand equals $$x^a(1-x)a^y\log a$$ for some $y\in(x,1)$ via mean value theorem and hence does not exceed $$x^a(1-x)a\log a\tag{1}$$ We can also use the fundamental inequality $e^x\geq 1+x$ for all real $x$ to get $$a-a^x=a\left(1-e^{(x-1)\log a}\right)\leq a(1-x)\log a$$ to reach the bound $(1)$ for the integrand (this is based on suggestion from robjohn in comments).
Upon integrating the expression $(1)$ on $[0,1]$ we get $$\frac{a\log a}{(a+1)(a+2)}$$ and it tends to $0$. It follows that integral at beginning of this answer tends to $0$ and the desired limit is thus $$\lim_{a\to\infty}a\int_0^1 x^a\,dx=1$$
A: Here is another method: Substituting $y = x^a$, or equivalent, $x = y^{1/a}$, the integral boils down to
\begin{align*}
I_a
:= \int_{0}^{1} a^x x^a \, \mathrm{d}x
&= \int_{0}^{1} a^{y^{1/a} - 1} y^{1/a} \, \mathrm{d}y \\
&= \int_{0}^{1} y^{1/a} \exp\bigl( (y^{1/a} - 1)\log a \bigr) \, \mathrm{d}y.
\end{align*}
Since the integrand is bounded between $0$ and $1$ (provided $a \geq 1$), by the dominated convergence theorem, this converges to:
\begin{align*}
\lim_{a\to\infty} I_a
&= \int_{0}^{1} \lim_{a \to \infty} y^{1/a} \exp\bigl( (y^{1/a} - 1)\log a \bigr) \, \mathrm{d}y \\
&= \int_{0}^{1} \, \mathrm{d}y
= 1.
\end{align*}
A: There has been a bit of activity on this question recently, so I considered an approach significantly different from my previous answer; different enough to warrant another answer rather than amending the existing answer.

Preliminary Limit
Let $u\in(0,1)$ and $a\ge1$. Then
$$
u^{\frac1{a+1}}\le1\tag1
$$
Furthermore, Bernoulli's Inequality gives
$$
u^{\frac1{a+1}}=(1+(1/u-1))^{-\frac1{a+1}}\ge1-\frac{1/u-1}{a+1}\tag2
$$
Thus,
$$
-\frac{1/u-1}{a+1}\le u^{\frac1{a+1}}-1\le0\tag3
$$
and since $a\ge1$,
$$
a^{-\frac{1/u-1}{a+1}}\le a^{u^{\frac1{a+1}}-1}\le1\tag4
$$
Therefore, by the fact that $\lim\limits_{a\to\infty}a^{\frac1{a+1}}=1$ and the Squeeze Theorem, we have
$$
\lim_{a\to\infty}a^{u^{\frac1{a+1}}-1}=1\tag5
$$

The Main Limit
$$
\begin{align}
\lim_{a\to \infty}\int_0^1a^xx^a\,\mathrm{d}x
&=\lim_{a\to \infty}\frac1{a+1}\int_0^1a^{u^{\frac1{a+1}}}\,\mathrm{d}u\tag{6a}\\
&=\lim_{a\to \infty}\frac{a}{a+1}\int_0^1a^{u^\frac1{a+1}-1}\,\mathrm{d}u\tag{6b}\\
&=\lim_{a\to \infty}\frac{a}{a+1}\ \lim_{a\to \infty}\int_0^1a^{u^\frac1{a+1}-1}\,\mathrm{d}u\tag{6c}\\
&=1\cdot\int_0^11\,\mathrm{d}u\tag{6d}\\[6pt]
&=1\tag{6e}
\end{align}
$$
Explanation:
$\text{(6a):}$ substitute $u=x^{a+1}$, then $x=u^{\frac1{a+1}}$ and $x^a\,\mathrm{d}x=\frac1{a+1}\,\mathrm{d}u$
$\text{(6b):}$ bring a factor of $a$ outside the integral
$\text{(6c):}$ the limit of a product is the product of the limits
$\text{(6d):}$ apply $(4)$ and $(5)$ and Dominated Convergence
$\text{(6e):}$ evaluate the integral
A: 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):

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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As $\ds{a \to \infty}$, the $\ds{a^{x}x^{a}}$ main contribution to the integral happens to be around $\ds{x = 1}$. Therefore,
\begin{align}
&\bbox[5px,#ddf]{%
\lim_{a \to \infty}\int_{0}^{1}a^{x}x^{a}\,\dd x} =
\lim_{a \to \infty}\int_{0}^{1}a^{1 - x}\,\,
\pars{1 - x}^{a}\,\dd x
\\[5mm] = & \
\lim_{a \to \infty}\int_{0}^{1}
\exp\pars{\pars{1 - x}\ln\pars{a} +
a\ln\pars{1 - x}}\,\dd x
\\[2mm] = & \
\lim_{a \to \infty}\,\,\,\overbrace{\int_{0}^{\infty}
\exp\pars{\ln\pars{a} - \bracks{a + \ln\pars{a}}x}
\,\dd x}^{\ds{Laplace's\ Method}} \
\\[5mm] = & 
\lim_{a \to \infty}\ {a \over a + \ln\pars{a}}
= \bbx{\large 1}
\end{align}
