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?
Best Answer
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!