Evaluate $\lim_{p\to 0+}\int_0^p t^{1+t} \mathrm{d}t$. Trying to integrate this is a pipe dream, so my hope is to somehow achieve the following scenario: $$L =\lim_{p\to 0+}\int_0^p g(t)\mathrm{d}t\leq \lim_{p\to 0+}\int_0^p t^{1+t} \mathrm{d}t\leq \lim_{p\to 0+}\int_0^p h(t) \mathrm{d}t = L $$
Question is: how should I go about finding $g, h$?
$t^{1+t} = e^{(1+t)\ln t}$. Considering the $e^y\geq e^x$ if $y\geq x$, then $(1+t)t\geq (1+t)\ln t$.
$e^{(1+t)\ln t}\leq e^{(1+t)t}$ should be true in our area of interest (approaching $0$ from the right).
Meanwhile I also noticed that $e^{f(x)}> 0$ for any $f,x$ so our lower boundary could be $0$.
What of the upper boundary? Is it also zero and if so, why? I can't follow through with my comparison attempt, would like clarification on that bit.
If $0 < t < p < 1$ then $0 < t^{1+t} < t$. Thus if $p < 1$ you have $$0 \le \int_0^p t^{1+t} \, dt \le \int_0^p t \, dt = \frac{p^2}{2}.$$ Now let $p \to 0^+$.