求积分

dim

已知p>0,求$\int_0^{\infty}e^{-px^2}\sin{x^2}dx$.先谢过了！

战巡

回复 1# dim


考虑下面这个积分：
\[\int_0^{+\infty}e^{-px^2}[i\sin(x^2)+\cos(x^2)]dx\]
\[=\int_0^{+\infty}e^{-px^2}e^{ix^2}dx=\frac{\sqrt{\pi}}{2}·\frac{1}{\sqrt{p-i}}\]
于是
\[\int_0^{+\infty}e^{-px^2}\sin(x^2)dx\]
就是上面这个玩意的虚部，具体如何求虚部我不写了，直接给结果
\[\int_0^{+\infty}e^{-px^2}\sin(x^2)dx=Im(\frac{\sqrt{\pi}}{2}·\frac{1}{\sqrt{p-i}})=\frac{\sqrt{\pi}}{2}·\frac{\sin(\frac{1}{2}\cot^{-1}(p))}{(1+p^2)^{\frac{1}{4}}}\]

dim

回复 2# 战巡


    大神有无初等一点的解法？复变现在还不太懂

青青子衿

回复  战巡
    大神有无初等一点的解法？复变现在还不太懂
dim 发表于 2016-4-8 00:30

\begin{align*}
I(k)&=\int_0^{+\infty}e^{-kx^2}\sin x^2{\rm\,d}x\\
I^2(k)&=\left(\int_0^{+\infty}e^{-kx^2}\sin x^2{\rm\,d}x\right)^2\\
&=\int_0^{+\infty}\int_0^{+\infty}e^{-k\left(x^2+y^2\right)}\sin x^2\sin y^2{\rm\,d}x{\rm\,d}y\\
&=\int_0^{+\infty}\int_0^{+\infty}e^{-k\left(x^2+y^2\right)}\cdot\dfrac{\cos\left(x^2-y^2\right)-\cos\left(x^2+y^2\right)}{2}{\rm\,d}x{\rm\,d}y\\
&=-\dfrac{1}{2}\int_0^{+\infty}\int_0^{+\infty}e^{-k\left(x^2+y^2\right)}\cdot\left[\cos\left(x^2+y^2\right)-\cos\left(x^2-y^2\right)\right]{\rm\,d}x{\rm\,d}y\\
&=-\dfrac{1}{2}\int_0^{\frac{\pi}{2}}\int_0^{+\infty}e^{-k\,r^2}\cdot\left[\cos\left(r^2\right)-\cos\left(r^2\cos2\theta\right)\right]r{\rm\,d}r{\rm\,d}\theta\\
&=-\dfrac{1}{\color{red}{4}}\int_0^{\frac{\pi}{2}}\int_0^{+\infty}e^{-k\,r^2}\cdot\left[\cos\left(r^2\right)-\cos\left(r^2\cos2\theta\right)\right]{\rm\,d}r^2{\rm\,d}\theta\\
&=-\dfrac{1}{4}\int_0^{\frac{\pi}{2}}\int_0^{+\infty}e^{-k\,u}\cdot\left[\cos\left(u\right)-\cos\left(u\cos2\theta\right)\right]{\rm\,d}u{\rm\,d}\theta\\
&=-\dfrac{1}{4}\int_0^{\frac{\pi}{2}}\left[\dfrac{k}{k^2+1}-\frac{k}{k^2+\left(\cos2\theta\right)^2}\right]{\rm\,d}\theta\\
&=-\dfrac{1}{4}\int_0^{\frac{\pi}{2}}\dfrac{k}{k^2+1}{\rm\,d}\theta+\dfrac{1}{4}\int_0^{\frac{\pi}{2}}\frac{k}{k^2+\left(\cos2\theta\right)^2}{\rm\,d}\theta\\
&=-\dfrac{1}{4}\cdot\frac{\pi}{2}\cdot\dfrac{k}{k^2+1}+\dfrac{1}{8}\int_0^{\frac{\pi}{2}}\frac{k}{k^2+\left(\cos2\theta\right)^2}{\rm\,d}(2\theta)\\
&=-\dfrac{1}{4}\cdot\frac{\pi}{2}\cdot\dfrac{k}{k^2+1}+\dfrac{k}{8}\int_0^{\pi}\frac{1}{k^2+\cos^2\varphi}{\rm\,d}\varphi\\
&=-\dfrac{k\pi}{8\left(k^2+1\right)}+\dfrac{k}{8}\int_0^{\pi}\frac{{\rm\,d}\varphi}{k^2\cos^2\varphi+k^2\sin^2\varphi+\cos^2\varphi}\\
&=-\dfrac{k\pi}{8\left(k^2+1\right)}+\dfrac{k}{8}\int_0^{\pi}\frac{{\rm\,d}\varphi}{\left(k^2+1\right)\cos^2\varphi+k^2\sin^2\varphi}\\
&=-\dfrac{k\pi}{8\left(k^2+1\right)}+\dfrac{k}{8}\int_0^{\pi}\frac{\frac{1}{\cos^2\varphi}{\rm\,d}\varphi}{\left(k^2+1\right)+k^2\frac{\sin^2\varphi}{\cos^2\varphi}}\\
&=-\dfrac{k\pi}{8\left(k^2+1\right)}+\dfrac{k}{8}\int_0^{\pi}\frac{\frac{1}{k^2+1}{\rm\,d}\tan\varphi}{1+\frac{k^2}{k^2+1}\tan^2\varphi}\\
&=-\dfrac{k\pi}{8\left(k^2+1\right)}+\dfrac{k}{8}\cdot\frac{1}{k\sqrt{k^2+1}}\int_0^{\pi}\frac{\frac{k}{\sqrt{k^2+1}}{\rm\,d}\tan\varphi}{1+\frac{k^2}{k^2+1}\tan^2\varphi}\\
&=-\dfrac{k\pi}{8\left(k^2+1\right)}+\frac{1}{8\sqrt{k^2+1}}\int_0^{\pi}\frac{{\rm\,d}\left(\frac{k}{\sqrt{k^2+1}}\tan\varphi\right)}{1+\frac{k^2}{k^2+1}\tan^2\varphi}\\
&=-\dfrac{k\pi}{8\left(k^2+1\right)}+\frac{\pi}{8\sqrt{k^2+1}}\\
&={\color{blue}{\frac{\pi\left(\sqrt{k^2+1}-k\right)}{8\left(k^2+1\right)}}}\\
I(k)&=\int_0^{+\infty}e^{-kx^2}\sin x^2{\rm\,d}x\\
&={\color{red}{\sqrt{\frac{\pi\left(\sqrt{k^2+1}-k\right)}{8\left(k^2+1\right)}}\,}}
\end{align*}
\(I(k)=\dfrac{\sqrt{\pi}\sin\left(\frac{\cot^{-1}k}{2}\right)}{2\sqrt[4]{1+k^2}}\)