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.

  • 1
    $\begingroup$ I guess the first step is to substitute $y = tx$. $\endgroup$
    – PhoemueX
    Jun 15 '18 at 23:00
  • $\begingroup$ Yes that was my first step. $\endgroup$ 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.

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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.

  • $\begingroup$ 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 $t\rightarrow(1-t)$. $\endgroup$
    – Diger
    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 $0<t_1<t_2\leq\frac{1}{2}$ $$ x\int_{t_1}^{t_2} e^{xt \log t} \, {\rm d}t \leq x\int_{t_1}^{t_2} e^{xt \log t_2} \, {\rm d}t = \frac{e^{xt_2 \log t_2}-e^{xt_1 \log t_2}}{\log t_2} $$ 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 $$ \lim_{x \rightarrow \infty} I \leq 1 \, . $$


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