Intuitive but hard question about an integral? Let $f \colon [0,1]\rightarrow \mathbb{R}$ be a function with continuous derivative such that $f(1)=1$. Evaluate $$\lim_{y\rightarrow \infty}\int_0^1yx^yf(x)dx$$
 A: All you need is that $f$ is continuous with $f(1) = 1$.  Note that
for any $\delta \in (0,1)$, 
$$\int_0^{1-\delta} y x^y\; dx = \dfrac{y}{y+1} (1-\delta)^{y+1} \to 0$$ 
while 
$$ \int_0^1 y x^y\; dx = \dfrac{y}{y+1}\to 1$$
Take $\delta$ so that $|f(x) - f(1)| < \epsilon$ for $1-\delta < x \le 1$, and
use the fact that $f$ is bounded... 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\fermi\pars{1} = 1}$.

$\large\tt\mbox{Heuristically !!!}:$

\begin{align}&\color{#66f}{\large%
\lim_{y\ \to\ \infty}\int_{0}^{1}yx^{y}\fermi\pars{x}\,\dd x}
=\lim_{y\ \to\ \infty}\int_{0}^{1}y\pars{1 - x}^{y}\fermi\pars{1 - x}\,\dd x
\\[5mm]&=\lim_{y\ \to\ \infty}\int_{0}^{1}
y\exp\pars{y\ln\pars{1 - x} + \ln\pars{\fermi\pars{1 - x}}}\,\dd x
\\[5mm]&=\lim_{y\ \to\ \infty}\int_{0}^{\infty}
y\exp\pars{-yx - \fermi'\pars{1}x}\,\dd x
=\lim_{y\ \to\ \infty}{y \over y + \fermi'\pars{1}}=\color{#66f}{\LARGE 1}
\end{align}
