反常积分∫ln(sin x)

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7_DefInt.pdf
\begin{equation}\int_0^{\pi/2}\ln\sin(x)\end{equation}
(1) 中的积分是反常积分，因为该函数在 $x=0$ 处是无穷大。在本段中，我们证明它收敛。
由于 $\sin x\ge 2x/\pi$ 对于 $0\le x\le\pi/2$，我们有
\begin{equation}
0 \ge \ln(\sin x) \ge \ln(2x/\pi)
\end{equation}
由于$\int_0^{π/2}\ln(2x/\pi)\,dx=\fracπ2\int_0^1\ln x\,dx= \fracπ2x(\ln x - 1)|_0^1= -\fracπ2$，通过比较检验法，(1) 收敛。

现在，我们使用 $\sin x = 2\sin(x/2)\cos(x/2)$ 和换元法来计算积分。首先
\begin{align}
\int_0^{\pi/2}\ln(\sin x)\;dx
&= \int_0^{\pi/2}\ln\Bigl(2\sin(x/2)\cos(x/2)\Bigr)\; dx\cr
&= \int_0^{\pi/2}\ln 2\;dx
+ \int_0^{\pi/2}\ln(\sin(x/2))\;dx
+ \int_0^{\pi/2}\ln(\cos(x/2))\;dx.
\end{align}
Now let $x/2=u$ in the last two integrals to obtain
$$
\int_0^{\pi/2}\ln(\sin x)\;dx
= (\pi\ln 2)/2
+ 2\int_0^{\pi/4}\ln(\sin u)\;du
+ 2\int_0^{\pi/4}\ln(\cos u)\;du.
$$
Now let $u=\pi/2-t$ in the last integral to obtain
$$
\int_0^{\pi/4}\ln(\cos u)\;du
= -\int_{\pi/2}^{\pi/4}\ln(\sin t)\;dt
= \int_{\pi/4}^{\pi/2}\ln(\sin t)\;dt,
$$
since $\cos(A-B)=\cos A\cos B + \sin A\sin B$.
Putting all this together we have
\begin{align}
\int_0^{\pi/2}\ln(\sin x)\;dx
&= (\pi\ln 2)/2
+ 2\int_0^{\pi/4}\ln(\sin u)\;du
+ 2\int_{\pi/4}^{\pi/2}\ln(\sin t)\;du\cr
&= (\pi\ln 2)/2 + 2\int_0^{\pi/2}\ln(\sin x)\;dx.\end{align}
Hence the value of the integral in (1) is $-(\pi\ln 2)/2$.

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