求极限
\[ \lim_{x\to+\infty}\sqrt{x}\int_{0}^{\frac{\pi}{4}}e^{x(\cos{t}-1)}\cos{t}dt \]

解
注意到
\[ \lim_{x\to 0}{\cos{x}}=1 \]
于是，对于任意$\varepsilon$,存在$\delta>0$,当$0<x<\delta$时
\[ \cos{x}>1-\varepsilon \]

\[ I=\int_{0}^{\delta}+\int_{\delta}^{\frac{\pi}{4}} \]

\[ \sqrt{x}\int_{0}^{\delta}e^{x(\cos{t}-1)}\cos{t}dt\leq  \sqrt{x}\int_{0}^{\delta}e^{x(\cos{t}-1)}dt\]
令$y=x(1-\cos{t}),\Rightarrow t=\arccos\left(1-\frac{y}{x}\right),\Rightarrow dt=\frac{1}{x\sqrt{1-\left(1-\frac{y}{x}\right)^{2}}}dy $,则
\begin{align*}
\sqrt{x}\int_{0}^{x(1-\cos\delta)}e^{x(\cos{t}-1)}dt&=\sqrt{x}\int_{0}^{x(1-\cos\delta)}e^{-y}\frac{1}{x\sqrt{1-\left(1-\frac{y}{x}\right)^{2}}}dy\\
&=\frac{1}{\sqrt{2}}\int_{0}^{x(1-\cos\delta)}e^{-y}y^{-\frac{1}{2}}\cdot \frac{1}{\sqrt{1-\frac{y}{2x}}}dy\\
&=\frac{1}{\sqrt{2}}\int_{0}^{x(1-\cos\delta)}e^{-y}y^{-\frac{1}{2}}\cdot\left(\frac{1}{\sqrt{1-\frac{y}{2x}}}-1\right)dy+\frac{1}{\sqrt{2}}\int_{0}^{x(1-\cos\delta)}e^{-y}y^{-\frac{1}{2}}dy\\
&=A+B
\end{align*}
显然有
\[ A=\frac{1}{\sqrt{2}}\int_{0}^{x(1-\cos\delta)}e^{-y}y^{-\frac{1}{2}}\cdot \frac{\frac{y}{2x}}{\sqrt{1-\frac{y}{2x}}\left(1+\sqrt{1-\frac{y}{2x}}\right)}dy\leq \frac{1}{2\sqrt{2}x}\int_{0}^{x}e^{-y}y^{\frac{1}{2}}dy\to 0   \]
\[ B=\frac{1}{\sqrt{2}}\int_{0}^{x(1-\cos\delta)}e^{-y}y^{-\frac{1}{2}}dy\to \sqrt{\frac{\pi}{2}}\qquad (x\to+\infty) \]

另外
\[ \sqrt{x}\int_{0}^{\delta}e^{x(\cos{t}-1)}\cos{t}dt\geq  \sqrt{x}\int_{0}^{\delta}e^{-\frac{1}{2}xt^{2}}(1-\varepsilon)dt\to \sqrt{\frac{\pi}{2}}(1-\varepsilon) \]
而不难证明
\[ \lim_{x\to+\infty}\int_{\delta}^{\frac{\pi}{4}}e^{x(\cos{t}-1)}\cos{t}dt=0 \]
这里只要用
\[ \cos{t}-1=-2\sin^{2}{\frac{t}{2}}\leq -2\left(\frac{t}{\pi}\right)^{2} \]
并注意到替换后的反常积分收敛就好，最后，由$\varepsilon$的任意性，得
\[ \lim_{x\to+\infty}\sqrt{x}\int_{0}^{\frac{\pi}{4}}e^{x(\cos{t}-1)}\cos{t}dt=\sqrt{\frac{\pi}{2}} \]