# 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.

• 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.
– robjohn
May 12 at 8:14
• You loose the argument when you say 'WLOG, let $d_a<1/2$'. May 12 at 8:25

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

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