证明 $\displaystyle\int_{0}^{+\infty}\frac{1}{(1+x^2)^{\frac{3}{2}}}\mathrm{d}x=1$.
由这个积分很容易推出
\[
\int_{0}^{+\infty}\frac{1}{(\rho^2+z^2)^{3/2}}\mathrm{d}z=\frac{1}{\rho^2}.
\]
这个积分很重要, 物理学中会用到.
1
首先说结论, 这个积分等于 1.
\[
\begin{split}
\int_0^{+\infty}\frac{1}{(1+x^2)\sqrt{1+x^2}}\mathrm{d}x&=\int_0^{+\infty}\frac{1}{\sqrt{1+x^2}}\mathrm{d}\arctan x=\frac{\arctan x}{\sqrt{1+x^2}}\biggr|_{0}^{+\infty}-\int_{0}^{+\infty}\arctan x\mathrm{d}\frac{1}{\sqrt{1+x^2}}\\
&=0-\int_{0}^{+\infty}\arctan x\cdot(-\frac{1}{2})(1+x^2)^{-\frac{3}{2}}\cdot 2x\mathrm{d}x\\
&=\int_{0}^{+\infty}\frac{x\arctan x}{(1+x^2)^{3/2}}\mathrm{d}x\\
&=\frac{1}{2}\int_{0}^{+\infty}\frac{\arctan x}{(1+x^2)^{3/2}}\mathrm{d}x^2\\
&\xlongequal[x=\sqrt{u-1}]{u=1+x^2}\frac{1}{2}\int_{1}^{+\infty}\frac{\arctan\sqrt{u-1}}{u^{3/2}}\mathrm{d}u\\
&=(-1)\int_{1}^{+\infty}\arctan\sqrt{u-1}\mathrm{d}\frac{1}{\sqrt{u}}\\
&=(-1)\biggl[\frac{1}{\sqrt{u}}\arctan\sqrt{u-1}\biggr|_{1}^{+\infty}-\int_{1}^{+\infty}\frac{1}{\sqrt{u}}\mathrm{d}\arctan\sqrt{u-1}\biggr]\\
&=(-1)\biggl[0-\int_{1}^{+\infty}\frac{1}{\sqrt{u}}\cdot\frac{1}{1+(u-1)}\cdot\frac{1}{2\sqrt{u-1}}\mathrm{d}u\biggr]\\
&=\frac{1}{2}\int_{1}^{+\infty}\frac{1}{u\sqrt{u^2-u}}\mathrm{d}u=\frac{1}{2}\int_{1}^{+\infty}\frac{1}{u\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}}\mathrm{d}u,
\end{split}
\]
令 $u-\frac{1}{2}=\frac{1}{2}\sec t$, $t\in[0,\frac{\pi}{2})$, 则 $u=\frac{1}{2}(1+\sec t)$, $\mathrm{d}u=\frac{1}{2}\sec t\cdot\tan t\mathrm{d}t$, $\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}=\frac{1}{2}\tan t$. 于是上面的积分等于
\[
\begin{split}
&\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{1}{2}(1+\sec t)\cdot\frac{1}{2}\tan t}\cdot\frac{1}{2}\sec t\cdot\tan t\mathrm{d}t\\
=&\int_{0}^{\frac{\pi}{2}}\frac{\sec t}{1+\sec t}\mathrm{d}t\\
=&\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos t+1}\mathrm{d}t\\
=&\int_{0}^{\frac{\pi}{2}}\frac{1-\cos t}{(1+\cos t)(1-\cos t)}\mathrm{d}t\\
=&\int_{0}^{\frac{\pi}{2}}\frac{1-\cos t}{\sin^2 t}\mathrm{d}t\\
=&\biggl(\frac{1-\cos t}{\sin t}\biggr)\biggr|_{0}^{\frac{\pi}{2}}\\
=&\frac{1-\cos\frac{\pi}{2}}{\sin\frac{\pi}{2}}-\lim_{t\rightarrow 0}\frac{1-\cos t}{\sin t}\\
=&1-\lim_{t\rightarrow 0}\frac{\sin t}{\cos t}\\
=&1.
\end{split}
\]