# Does anyone know a simple proof of $\lim_{x\rightarrow\infty} x\int_0^1 t^{x t} dt = 1$

I proved the identity: $$\lim_{x\rightarrow\infty} x\int_0^1 t^{x t} dt = 1$$ but my proof is very long and complicated. I'm wondering if there is a simpler way to do it.

Putting a lower bound of 1 is pretty straightforward, because $$t^{xt}\ge t^x$$ for $$t\in[0,1]$$: $$\lim_{x\rightarrow\infty} x\int_0^1 t^{x t} dt \ge \lim_{x\rightarrow\infty} x\int_0^1 t^{x} dt = \lim_{x\rightarrow\infty} \frac{x}{x+1} = 1$$ I did manage to prove an upper bound of 1 by taking Riemann sums after substituting $$y = tx$$ but I am wondering if there is a more elegant way to do it.

• I guess the first step is to substitute $y = tx$. Jun 15 '18 at 23:00
• Yes that was my first step. Jun 16 '18 at 23:16

With integration by parts one can prove for $$x \to \infty$$ the asymptotic formula below. The condition for $$h(t)$$ is that it is increasing on (0,1) and $$v$$ is assumed to be >0 and bounded. (One may also insert into the integrand a continuous function with certain growth conditions, but I'll leave that generalization aside.)

$$\int_0^1 \exp{(x \,h(t) )} (1-t)^{v-1} \, dt \sim \frac{\Gamma(v)}{x^v} \frac{\exp{(x h(1) )}}{(h'(1))^v}$$

An earlier answer of mine used this but I was sloppy and used it even though the integrand is not increasing over all of (0,1). Fortunately there is a way to salvage the argument. The picture illustrates that for large $$x$$ only the regions around the endpoints of $$t=0$$ and $$t=1$$ are important. Split the integrand at $$t=1/e,$$ the minimum of $$t \, \log{t}.$$ Furthermore write $$\exp(x t\,\log{t} ) = \exp{(-x/e)} \exp{\big(x(t\log{t} + 1/e})\big)= \exp{(-x/e)} \exp{\big(x\, h_1(t)\big)}$$ This is done because $$h_1(t) \ge 0$$ for t in (0,1). The split means $$I(x)=\int_0^1 \exp{x(\,t\,\log{t})} dt =\exp{(-x/e)}\Big( \underbrace{\int_0^{1/e} \exp{\big(x\, h_1(t)\big)} \,dt}_{I_1} + \underbrace{\int_{1/e}^{1} \exp{\big(x\, h_1(t)\big)} \,dt }_{I_2} \Big)$$ $$I_2$$ is easy to deal with from the formula on the top of the page. The fact that the lower limit does not extend to 0 can be handled with this argument: We could augment $$h_1$$ with a function that has a value <0 over (0,1/e) and the criterion of non-decreasing would still be satisfied, and the integral over (0,1/e) would be exponentially smaller than the part we are interested in. Using the formula we find $$h(1)=1/e,$$ $$h'(1) = 1$$ and thus $$\exp{(-x/e)}\, I_2 = \exp{(-x/e)}\, \cdot \frac{ \exp{(x/e)}}{x} \sim \frac{1}{x}$$ Now we want to show $$\exp{(-x/e)}I_1$$ is exponentially smaller. Rescale the integral to find $$I_1 = \frac{1}{e}\, \int_0^1 \exp{(x/e \,t\log{(t/e)}) }\, dt= \frac{1}{e}\, \int_0^1 \exp{\big(x/e \,t(\log{t} - 1)\big) }\,dt$$ $$= \frac{1}{e}\, \int_0^1 \exp{\big(x/e \,(1-t)(\log{(1-t)} - 1)\big) }\,dt =$$ $$=\frac{ \exp{(-x/e)}}{e} \int_0^1 \exp{\big(x/e\big(\underbrace{1+ \,(1-t)(\log{(1-t)} - 1)}_{h_2(t)}\big)\big) }\,dt$$ From the first line to the second the substitution $$t \to 1-t$$ was performed. In the last line of the previous calculation $$h_2$$ has been defined so that it is $$\ge 0$$ on (0,1). If we try to use the formula at the top of page then we find $$h_2'(1) = \infty,$$ and that is not allowed so the formula is not directly applicable. However, we only want a bound so define $$h_3(t) = \frac{t^2}{2}(1+t) \ge h_2(t) , \text{ so } h_3'(1) = 5/2, h_3(1)=1.$$ We can thus use the formula and find $$I_1 < \frac{ \exp{(-x/e)}}{e} \cdot \frac{ \exp{(x/e)}}{5/2 x} = \frac{2}{5 e x}$$ Thus indeed $$\exp{(-x/e)} I_1$$ is exponentially smaller.

• Am I right that $$I_1 =\frac{1}{e}\, \int_0^1 \exp{(x/e \,t\log{(t/e)}) }\, dt$$ originates from the full $t\log t$ exponent and not from $I_1$ defined via $h_1$ above? Since then you are multiplying with $e^{-x/e}$ one times too much and your actual estimate $2/(5ex)$ is not exponentially smaller. The integral from $0$ to $1/e$ indeed vanishes, but this is precisely because $(t\log t)'=1+\log t$ goes to $-\infty$ for $t\rightarrow 0$. Any finite slope at $0$ would yield a finite result, which happens to be the case for your function $h_3$ about $t=1$ (after your inversion $t\rightarrow(1-t)$. Oct 24 '18 at 16:48

I would first substitute $$t={\rm e}^{-u}$$ to obtain $$x\int_{0}^{\infty }{{\rm e}^{-u \left( x{{\rm e}^{-u}}+1 \right) }} \,{\rm d}u$$ and split the integral $$x\int_{0}^{\frac{1}{\sqrt{x}}}{{\rm e}^{-u \left( x{{\rm e}^{-u}}+1 \right) }} \,{\rm d}u + x\int_{\frac{1}{\sqrt{x}}}^\infty{{\rm e}^{-u \left( x{{\rm e}^{-u}}+1 \right) }} \,{\rm d}u \, .$$

For $$x\rightarrow \infty$$ you can replace $${\rm e}^{-u}$$ by $$1$$ in the first term and carry out the integral which yields $$1+{\cal O}\left(\frac{1}{x}\right)$$. The second integral can be shown to be $${\cal O}\left(\frac{1}{x}\right)$$.

For a more rigorous proof you can sandwich the first term with $${\rm e}^{-u} \geq 1-u$$ and $${\rm e}^{-u}\leq 1-\sqrt{x}\left(1-{\rm e}^{-\frac{1}{\sqrt{x}}}\right) u$$ and use the series expansion of the arising error function.

Another argument goes as follows: Again split the integral

$$\underbrace{x\int_0^1 e^{xt\log t} \, {\rm d}t}_{I} = \underbrace{x\int_0^{1/2} e^{xt\log t} \, {\rm d}t}_{I_1} + \underbrace{x\int_{1/2}^1 e^{xt\log t} \, {\rm d}t}_{I_2} \, .$$ Then for any $$0 $$x\int_{t_1}^{t_2} e^{xt \log t} \, {\rm d}t \leq x\int_{t_1}^{t_2} e^{xt \log t_2} \, {\rm d}t = \frac{e^{xt_2 \log t_2}-e^{xt_1 \log t_2}}{\log t_2}$$ which vanishes for $$x\rightarrow \infty$$. Thus for any $$\delta > 0$$ we can choose $$\epsilon > 0$$ such that $$x\int_0^\epsilon e^{xt \log t} \, {\rm d}t \leq x\int_0^\epsilon e^{xt \log \epsilon} \, {\rm d}t = \frac{e^{x\epsilon \log \epsilon} - 1}{\log \epsilon} \leq -\frac{1}{\log \epsilon} < \delta$$ for $$x\epsilon >> 1$$ and so $$\lim_{x \rightarrow \infty} I_1 = 0 \, .$$

On the other hand since $$t\leq e^{t-1}$$ for all $$1/2 \leq t \leq 1$$ we have $$I_2 \leq x\int_{1/2}^1 e^{(t-1)xt} \, {\rm d}t = -\frac{i}{2} \, \sqrt{\pi x} \, e^{-x/4} \, {\rm erf}\left(\frac{i}{2}\,\sqrt{x}\right) = 1 + \frac{2}{x} + {\cal O}\left(x^{-2}\right)$$ and therefore $$\lim_{x \rightarrow \infty} I \leq 1 \, .$$