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发表于 2020-10-30 01:31
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本帖最后由 elim 于 2020-10-30 09:14 编辑
题:试证\(\,f(x)=\displaystyle{\small\int_0^{\infty}}{\large e^{-(t-\frac{x}{t})^2}}dt = {\small\frac{\sqrt{\pi}}{2}}e^{-2(|x|-x)}\).
证:定义\(\,f(x):=\displaystyle{\small\int_0^{\infty}}e^{-(t-\frac{x}{t})^2}dt.\quad\)则对\(\,a> 0\) 有
\(\;\displaystyle f(a)\,\overset{u=\frac{a}{t}}{=\hspace{-3px}=}{\small\int_0^{\infty}\frac{a}{u^2}}e^{-(u-\frac{a}{u})^2}du\overset{\small\color{red}{(^*)}}{=}{\scriptsize\frac{1}{2}}{\small\int_0^{\infty}}({\small 1+\frac{a}{u^2}})e^{-(u-\frac{a}{u})^2}du\)
\(\qquad\displaystyle\overset{v=u-\frac{a}{u}}{=\hspace{-3px}=}{\scriptsize\frac{1}{2}}{\small\int_0^{\infty}}e^{-v^2}dv=f(0)={\small\frac{\sqrt{\pi}}{2}}.\)
\(\quad\)进而由\(\underset{\,}{\,}(t-\frac{x}{t})^2=(t-\frac{|x|}{t})^2\small+2(|x|-x)\) 得
\(\qquad\qquad\boxed{\;{\small\int_0^{\infty}}e^{-(t-\frac{x}{t})^2}dt={\scriptsize\frac{\sqrt{\pi}}{2}}e^{-2(|x|-x)}.\;}\)
\(\small\color{red}{(^*)}\;\;\color{grey}{f(a)}\,\)的积分变形与积分定义的算术平均.
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