7
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ You cannot say that $d_a\lt1/2$ and that $\left|a-a^x\right|\lt\epsilon$ for all $x\in[1/2,1]$ since $a$ can be arbitrarily large. $\endgroup$
    – robjohn
    May 12 at 8:14
  • 2
    $\begingroup$ You loose the argument when you say 'WLOG, let $d_a<1/2$'. $\endgroup$ May 12 at 8:25

2 Answers 2

11
$\begingroup$

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


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, $$ \lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x=1\tag4 $$


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 $$ \lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x=1\tag8 $$

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.