Prove $\lim\limits_{x \to +\infty}\int_0^{\pi} xe^{-x\sin t}{\rm d}t=2$. Someone gives a proof as follows：

Above all, notice that \begin{align*} I(x):&=\int_0^{\pi} x{\rm  e}^{-x\sin t}{\rm d}t=\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\sin t}{\rm  d}t+\overbrace{\int_{\frac{\pi}{2}}^{\pi} x{\rm e}^{-x\sin t}{\rm  d}t}^{t~ \mapsto ~t+\frac{\pi}{2}}\\ &=\int_0^{\frac{\pi}{2}} x{\rm  e}^{-x\sin t}{\rm d}t+\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\sin  \left(t+\frac{\pi}{2}\right)}{\rm d}t\\ &=\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\sin t}{\rm d}t+\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\cos t}{\rm  d}t\\ &=\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\sin t}{\rm  d}t+\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\cos\left(\frac{\pi}{2}-  t\right)}{\rm d}t\\ &=\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\sin t}{\rm
d}t+\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\sin t}{\rm d}t\\
&=2\int_0^{\frac{\pi}{2}} x{\rm e}^{-x\sin t}{\rm d}t. \end{align*}
  Consider making a substitution that $\theta:=tx.$ Then $x=\theta/t,
 {\rm d}\theta=x{\rm d}t.$ Thus $$I(x):=2\int_0^{\frac{\pi x}{2}}
 \exp\left(-x\sin\frac{\theta}{x}\right){\rm d}\theta.$$ Since
   $\theta/x \in[0,\pi/2],$ and $f(x):=\dfrac{\sin x}{x}$ decreases over
   $(0,\pi/2]$, hence $ \sin \dfrac{\theta}{x}\ge \dfrac{2\theta}{\pi x}.$
  Therefore $ -x\sin \dfrac{\theta}{x}\le -\dfrac{2\theta}{\pi },$ and
   further we obtain $$\left|
\exp\left(-x\sin\frac{\theta}{x}\right)\right|=
 \exp\left(-x\sin\frac{\theta}{x}\right)\le \rm
 e^{-\frac{2\theta}{\pi}},$$ the right hand side of which is integrable
   over $ [0,+\infty) .$ Moreover $$\lim_{x \to
 +\infty}\exp\left(-x\sin\frac{\theta}{x}\right)=\exp\lim_{x \to +\infty}\left(\frac{\sin \frac{\theta}{x}}{\frac{\theta}{x}}\cdot -\theta\right)=e^{-\theta}.$$ Now, we can exchange the orders of the limit and the integral by Lebesgue dominated convergence theorem and
   obtain $$\lim_{x \to +\infty}I(x)=2\int_0^{+\infty}{\rm
 e}^{-\theta}{\rm d}\theta=2.$$

Is this correct? I don't know Lebesgue dominated convergence theorem well. Is there another proof more elementary?
 A: 
The proof in the OP is a standard way forward when equiped with the Dominated Convergence Theorem.  Herein, we present a way forward that relies on elementary calculus only and uses integration by parts.  To that end, we now proceed.


Let $I(x)$ be given by the integral 
$$\begin{align}
I(x)&=\int_0^\pi xe^{-x\sin(t)}\,dt\\\\
&=2\int_0^{\pi/2} xe^{-x\sin(t)}\,dt\\\\
&=2\int_0^1 xe^{-x\sin(t)}\,dt+2\int_1^{\pi/2} xe^{-x\sin(t)}\,dt\tag1
\end{align}$$

As $x\to \infty$, it is evident that the second integral on the right-hand side of $(1)$ approaches $0$ (since $\lim_{x\to\infty}xe^{-x\sin(1)}=0$).  

For the first integral, we integrate by parts with $u=\sec(t)$ and $v=-\frac1x e^{-x\sin(t)}\cos(t)$.  Proceeding we find that 
$$\begin{align}
2\int_0^1 xe^{-x\sin(t)}\,dt&=2\left.\left(- \sec(t)e^{-x\sin(t)}\right)\right|_0^1+2\int_0^1 \sec(t)\tan(t)e^{-x\sin(t)}\,dt\\\\
&=2-2\sec(1)e^{-x\sin(1)}+2\int_0^1 \sec(t)\tan(t)e^{-x\sin(t)}\,dt\tag2
\end{align}$$
Letting $x\to\infty$ in $(2)$ yields the coveted limit
$$\lim_{x\to\infty}\int_0^\pi xe^{-x\sin(t)}\,dt=2$$
as was to be shown!
A: Let $ x>1 $ :
First of all, as you've shown : \begin{aligned}\int_{0}^{\pi}{x\,\mathrm{e}^{-x\sin{t}}\,\mathrm{d}t}=\int_{0}^{\frac{\pi}{2}}{x\,\mathrm{e}^{-x\sin{t}}\,\mathrm{d}t}+\int_{\frac{\pi}{2}}^{\pi}{x\,\mathrm{e}^{-x\sin{t}}\,\mathrm{d}t}\end{aligned}
Substituting $ t=\pi-y $ in the second term, we get : \begin{aligned}\int_{0}^{\pi}{x\,\mathrm{e}^{-x\sin{t}}\,\mathrm{d}t}=2\int_{0}^{\frac{\pi}{2}}{x\,\mathrm{e}^{-x\sin{t}}\,\mathrm{d}t}\end{aligned}
Now substituting $ \small\left\lbrace\begin{aligned}y&=x\sin{t}\\ \mathrm{d}t&=\frac{\mathrm{d}y}{\sqrt{1-\frac{y^{2}}{x^{2}}}}\end{aligned}\right. $, we get : \begin{aligned}\int_{0}^{\pi}{x\,\mathrm{e}^{-x\sin{t}}\,\mathrm{d}t}=2\int_{0}^{x}{\frac{\mathrm{e}^{-y}}{\sqrt{1-\frac{y^{2}}{x^{2}}}}\,\mathrm{d}y}\end{aligned}
We have : \begin{aligned}\left|\int_{0}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}-\int_{0}^{x}{\frac{\mathrm{e}^{-y}}{\sqrt{1-\frac{y^{2}}{x^{2}}}}\,\mathrm{d}y}\right|&=\left|\int_{x}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}-\int_{0}^{x}{\mathrm{e}^{-y}\left(1-\frac{1}{\sqrt{1-\frac{y^{2}}{x^{2}}}}\right)\mathrm{d}y}\right|\\ &=\left|\int_{x}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}-\frac{1}{x^{2}}\int_{0}^{x}{\frac{y^{2}\,\mathrm{e}^{-y}}{\sqrt{1-\frac{y^{2}}{x^{2}}}\left(1+\sqrt{1-\frac{y^{2}}{x^{2}}}\right)}\,\mathrm{d}y}\right|\\ &\leq\int_{x}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}+\frac{1}{x^{2}}\int_{0}^{x}{\frac{y^{2}\,\mathrm{e}^{-y}}{\sqrt{1-\frac{y^{2}}{x^{2}}}\left(1+\sqrt{1-\frac{y^{2}}{x^{2}}}\right)}\,\mathrm{d}y}\\ &\leq\int_{x}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}+\frac{1}{x^{2}}\int_{0}^{x}{\frac{y^{2}\,\mathrm{e}^{-y}}{\sqrt{1-\frac{y^{2}}{x^{2}}}}\,\mathrm{d}y}\\ &\leq\int_{x}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}+\frac{1}{x^{2}}\int_{0}^{x}{\frac{y^{2}}{\left(1+\frac{y^{2}}{2}\right)\sqrt{1-\frac{y^{2}}{x^{2}}}}\,\mathrm{d}y}\end{aligned}
(We used the fact that $ \mathrm{e}^{y}\geq 1+y+\frac{y^{2}}{2}\geq 1+\frac{y^{2}}{2} $ then we took its inverse)
Now since the integral we got can be computed, $ \int_{0}^{x}{\frac{y^{2}}{\left(1+\frac{y^{2}}{2}\right)\sqrt{1-\frac{y^{2}}{x^{2}}}}\,\mathrm{d}y}=\pi x\left(1-\frac{1}{\sqrt{1+\frac{x^{2}}{2}}}\right) $, we get that : $$ \left|\int_{0}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}-\int_{0}^{x}{\frac{\mathrm{e}^{-y}}{\sqrt{1-\frac{y^{2}}{x^{2}}}}\,\mathrm{d}y}\right|\leq\int_{x}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}+\frac{\pi}{x}\left(1-\frac{1}{\sqrt{1+\frac{x^{2}}{2}}}\right)\underset{x\to +\infty}{\longrightarrow}0 $$
Which means $$ \int_{0}^{x}{\frac{\mathrm{e}^{-y}}{\sqrt{1-\frac{y^{2}}{x^{2}}}}\,\mathrm{d}y}\underset{x\to +\infty}{\longrightarrow}\int_{0}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}=1 $$
Hence $$ \int_{0}^{\pi}{x\,\mathrm{e}^{-x\sin{t}}\,\mathrm{d}t}\underset{x\to +\infty}{\longrightarrow}2 $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\lim_{x \to \infty}\int_{0}^{\pi}x\expo{-x\sin\pars{t}}\dd t & =
\lim_{x \to \infty}\int_{-\pi/2}^{\pi/2}x\expo{-x\cos\pars{t}}\dd t =
2\lim_{x \to \infty}\int_{0}^{\pi/2}x\expo{-x\cos\pars{t}}\dd t
\\[5mm] & =
2\lim_{x \to \infty}\int_{0}^{\pi/2}x\expo{-x\sin\pars{t}}\dd t =
2\lim_{x \to \infty}\int_{0}^{\infty}x\expo{-xt}\dd t
\label{1}\tag{1}
\\[5mm] & =
2\lim_{x \to \infty}\int_{0}^{\infty}\expo{-t}\dd t = \bbx{\large 2}
\end{align}
In line (\ref{1}), I used the
Laplace Method.
