问题及解答

求 \[\int_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx.\]

[相关的积分]

证明:

\[
\int_0^1 \frac{1}{\sqrt{1-x^4}}dx\cdot\int_0^1 \frac{x^2}{\sqrt{1-x^4}}dx=\frac{\pi}{4}.
\]

如果我们记 $f(x)=\frac{1}{\sqrt{1-x^4}}$, $g(x)=\frac{x^2}{\sqrt{1-x^4}}$, 则

\[
\begin{aligned}
​f(x)+g(x)=\frac{1+x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1+x^2}{1-x^2}},\\
f(x)-g(x)=\frac{1-x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1-x^2}{1+x^2}},\\
​\end{aligned}
\]

于是 $(f(x)+g(x))(f(x)-g(x))=1$, 因此原积分

\[
​\begin{split}
\int_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx&=\int_0^1\sqrt{\frac{1-x^2}{1+x^2}}\\
​&=\int_0^1(f(x)-g(x))dx\\
&=\int_0^1 \frac{1}{\sqrt{1-x^4}}dx-\int_0^1 \frac{x^2}{\sqrt{1-x^4}}dx
​\end{split}
\]

References:

吉米多维奇, 《数学分析习题集题解》(五) 3872.

[分析]

设 $x^2=\sin\theta$, $\theta\in[0,\frac{\pi}{2}]$, 则 $x=\sqrt{\sin\theta}$, $dx=\frac{1}{2\sqrt{\sin\theta}}\cos\theta d\theta$.

\[
\begin{split}
\text{原式}&=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\theta}}{1+\sin\theta}\cdot\frac{\cos\theta}{2\sqrt{\sin\theta}}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2\theta}{2\sqrt{\sin\theta}(1+\sin\theta)}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\sin^2\theta}{2\sqrt{\sin\theta}(1+\sin\theta)}d\theta\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\sin\theta}{2\sqrt{\sin\theta}}d\theta\\
&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin\theta}}d\theta-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\theta}d\theta,
\end{split}
\]

其中积分

\[
\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\theta}d\theta\stackrel{t=\sqrt{\sin\theta}}{=}2\int_0^1\frac{t^2}{\sqrt{1-t^4}}dt
\]

是一个椭圆积分, 可以参考问题998 .

我们先将 $\int_0^1 x^{p-1}(1-x^m)^{q-1}dx$ 类型的定积分表示为 B-函数

令 $u=x^m$, 则 $x=u^{\frac{1}{m}}$, $dx=\frac{1}{m}u^{\frac{1}{m}-1}du$, 于是

\[
\begin{split}
​\int_0^1 x^{p-1}(1-x^m)^{q-1}dx &=\int_0^1 (u^{\frac{1}{m}})^{p-1}(1-u)^{q-1}\cdot\frac{1}{m}u^{\frac{1}{m}-1}du\\
​&=\frac{1}{m}\int_0^1 u^{\frac{p}{m}-1}(1-u)^{q-1}du\\
​&=\frac{1}{m}B(\frac{p}{m},q)\\
​&=\frac{1}{m}\frac{\Gamma(\frac{p}{m})\Gamma(q)}{\Gamma(\frac{p}{m}+q)}
​\end{split}
\]

上面最后一个等号用到了 B-函数与 $\Gamma$-函数之间的关系(见问题714).

于是

\[
\int_0^1\frac{1}{\sqrt{1-x^4}}dx=\int_0^1 x^{1-1}(1-x^4)^{\frac{1}{2}-1}dx=\frac{1}{4}\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{4}+\frac{1}{2})}=\frac{1}{4}\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{4})},
\]

\[
\int_0^1\frac{x^2}{\sqrt{1-x^4}}dx=\int_0^1 x^{3-1}(1-x^4)^{\frac{1}{2}-1}dx=\frac{1}{4}\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{4}+\frac{1}{2})}=\frac{1}{4}\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{5}{4})}.
\]

因此

\[
\begin{split}
&\int_0^1\frac{1}{\sqrt{1-x^4}}dx-\int_0^1\frac{x^2}{\sqrt{1-x^4}}dx\\
​=&\frac{1}{4}\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{4})}-\frac{1}{4}\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{5}{4})}\\
​=&\frac{1}{4}\Gamma(\frac{1}{2})\cdot\biggl[\frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}-\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})}\biggr]\\
​=&\frac{1}{4}\Gamma(\frac{1}{2})\cdot\biggl[\frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}-\frac{\Gamma(\frac{3}{4})}{\frac{1}{4}\Gamma(\frac{1}{4})}\biggr]\\
​=&\frac{1}{4}\Gamma(\frac{1}{2})\frac{\Gamma(\frac{1}{4})^2-4\Gamma(\frac{3}{4})^2}{\Gamma(\frac{1}{4})\Gamma(\frac{3}{4})}\\
​=&\frac{1}{4}\cdot\sqrt{\pi}\cdot\frac{\Gamma(\frac{1}{4})^2-4\Gamma(\frac{3}{4})^2}{\sqrt{2}\pi}\\
​=&\frac{\Gamma(\frac{1}{4})^2-4\Gamma(\frac{3}{4})^2}{4\sqrt{2\pi}}.
​\end{split}
\]

这里用到了公式 (见问题714)

\[
\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)},
\]

由此可得 $\Gamma(\frac{1}{2})=\sqrt{\pi}$, 以及 $\Gamma(\frac{1}{4})\Gamma(\frac{3}{4})=\sqrt{2}\pi$.