The Limit of a Wallis-like Integral Let
$\displaystyle I_n=\int^{\pi}_{0} e^x \sin^n x dx$,
then, what number does
$\displaystyle \sqrt{n} I_n$
tend to?
According to wolfram alpha it tends to 12.05... but I want to know what is this number.
I posted similarly titled question about two months ago: The limit and Asymptotic Behavior Wallis-like Integral but the theme here is not that.
 A: A key idea here is that the kernel
$$K_n(x) = \sqrt{n}\sin^n (x), \qquad 0 \leq x \leq \pi$$
is an approximation-to-the-identity (up to constant multiple). More precisely, for each continuous function $\varphi$ on $[0, \pi]$, the following is true:
$$ \lim_{n\to\infty} \int_{0}^{\pi} \varphi(x) K_n(x) \, \mathrm{d}x = \sqrt{2\pi} \phi\Bigl(\frac{\pi}{2}\Bigr) $$
Here is a proof:
\begin{align*}
\int_{0}^{\pi} \varphi(x) K_n(x) \, \mathrm{d}x
&= \int_{-\frac{\pi}{2}\sqrt{n}}^{\frac{\pi}{2}\sqrt{n}} \phi\Bigl( \frac{\pi}{2} + \frac{u}{\sqrt{n}} \Bigr) \cos^n(u/\sqrt{n}) \, \mathrm{d}u \tag{$\textstyle x = \frac{\pi}{2} + \frac{u}{\sqrt{n}}$}.
\end{align*}
Now by noting that $ \log(\cos t) \leq - \frac{t^2}{2} $ holds for all $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we have
$$ \cos^n(u/\sqrt{n}) \leq e^{-u^2/2}. $$
Moreover, using the asymptotic formula $\log(\cos t) = -\frac{t^2}{2}(1 + o(1))$ as $t \to 0$, it follows that for each fixed $u$ in the domain of integration,
$$ \cos^n(u/\sqrt{n}) \xrightarrow[n\to\infty]{} e^{-u^2/2}. $$
So by the dominated convergence theorem,
\begin{align*}
&\lim_{n\to\infty} \int_{-\frac{\pi}{2}\sqrt{n}}^{\frac{\pi}{2}\sqrt{n}} \phi\Bigl( \frac{\pi}{2} + \frac{u}{\sqrt{n}} \Bigr) \cos^n(u/\sqrt{n}) \, \mathrm{d}u \\
&= \int_{-\infty}^{\infty} \lim_{n\to\infty} \phi\Bigl( \frac{\pi}{2} + \frac{u}{\sqrt{n}} \Bigr) \cos^n(u/\sqrt{n}) \mathbf{1}_{\{ -\frac{\pi}{2}\sqrt{n} < u < \frac{\pi}{2}\sqrt{n}\}} \, \mathrm{d}u \\
&= \int_{-\infty}^{\infty} \phi\Bigl( \frac{\pi}{2} \Bigr) e^{-u^2/2} \, \mathrm{d}u \\
&= \sqrt{2\pi} \phi\Bigl( \frac{\pi}{2} \Bigr).
\end{align*}
As a corollary,
\begin{align*}
\lim_{n\to\infty} \int_{0}^{\pi} e^x K_n(x) \, \mathrm{d}x
= \sqrt{2\pi} e^{\pi/2}
\approx 12.05807862.
\end{align*}
A: Denoting $ \left(\forall n\in\mathbb{N}\right),\ I_{n}=\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x} $.
Let $ n \in\mathbb{N} :$
Integrating by parts : \begin{aligned}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x}=-\frac{n-1}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\left(\sin{x}-\cos{x}\right)\sin^{n-2}{x}\cos{x}\,\mathrm{d}x}\end{aligned} Reintegrating by parts : \begin{aligned} \int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x}=\frac{\left(n-1\right)^{2}}{4}\int_{0}^{\pi}{\mathrm{e}^{x}\left(\sin{x}+\cos{x}\right)\sin^{n-2}{x}\cos{x}\,\mathrm{d}x}+\frac{n-1}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\cos^{2}{x}\,\mathrm{d}x}\end{aligned}
Thus : \begin{aligned}\left(1+\frac{n-1}{2}\right)\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x}&=\frac{\left(n-1\right)^{2}}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\cos^{2}{x}\,\mathrm{d}x}+\frac{n-1}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\cos^{2}{x}\,\mathrm{d}x}\\ &=\frac{n\left(n-1\right)}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\left(1-\sin^{2}{x}\right)\mathrm{d}x}\\ \frac{n+1}{2}I_{n}&=\frac{n\left(n-1\right)}{2}\left(I_{n-2}-I_{n}\right)\end{aligned}
Thus : $$ \fbox{$\begin{array}{rcl}\left(1+n^{2}\right)I_{n}=n\left(n-1\right)I_{n-2}\end{array}$} $$
Define a sequence $ \left(u_{n}\right)_{n\geq 1} $ as follows : $$ \left(\forall n\in\mathbb{N}^{*}\right),\ u_{n}=\frac{n}{\left(n!\right)^{2}}\prod_{k=1}^{n}{\left(1+k^{2}\right)}I_{n-1}I_{n} $$
Since $ \left(u_{n}\right) $ is a constant sequence and $ \frac{I_{n+1}}{I_{n}}\underset{n\to +\infty}{\longrightarrow}1 $, we get : $$ I_{n}\underset{n\to +\infty}{\sim}\frac{n!\sqrt{2\,\mathrm{e}^{\pi}\sinh{\pi}}}{\sqrt{n\prod\limits_{k=1}^{n}{\left(1+k^{2}\right)}}} $$
We have $ \prod\limits_{k=1}^{n}{\left(1+k^{2}\right)}=\frac{\sinh{\pi}\,\Gamma\left(n+1-\mathrm{i}\right)\,\Gamma\left(n+1+\mathrm{i}\right)}{\pi}\underset{n\to +\infty}{\sim}\frac{\sinh{\pi}\,\left(n!\right)^{2}}{\pi} $, which means : $$\fbox{$\begin{array}{rcl} I_{n}\underset{n\to +\infty}{\sim}\sqrt{\frac{2\pi\,\mathrm{e}^{\pi}}{n}} \end{array}$} $$
The limit would, thus, be equal to : $$ \sqrt{2\pi\,\mathrm{e}^{\pi}} $$
A: There is a reduction formula for the antiderivative. Have a look at equation $2.662.1$ in Table of Integrals, Series, and Products (Seventh Edition)  by I.S. Gradshteyn and I.M. Ryzhik.
If you enjoy hypergeometric functions, there is a formal solution. Using the bounds, if I am not mistaken
$$I_n=-\frac{i  e^{\frac{i \pi  n}{2}} \left(e^{\pi(1+ni)}-1\right) \Gamma
   \left(-\frac{n+i}{2}\right) \Gamma (n+1)}{2^{n+1}\Gamma \left(\frac{n-i}{2}+1\right)}$$
How to take the limit or the asymptotics of the monster ???
Computing it for $n=10^k$, some results (waiting for an answer !)
$$\left(
\begin{array}{cc}
k & \text{result} \\
 1 & 12.30548841 \\
 2 & 12.08766068 \\
 3 & 12.06108749 \\
 4 & 12.05838001 \\
 5 & 12.05810876 \\
 6 & 12.05808163
\end{array}
\right)$$
