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.
 A: 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.
A: 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
$$
\lim_{x \rightarrow \infty} I \leq 1 \, .
$$
