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Putnam 2005 A5. $\displaystyle\int_{0}^{1} \frac{\ln (x+1)}{x^{2}+1} d x$
设 $\displaystyle I(a)=\int_{0}^{1} \frac{\ln (a x+1)}{x^{2}+1} d x$
这意味着我们的原始积分是 $I(1)$ 并且 $I(0) = 0$。
\begin{aligned} \frac{\partial I}{\partial a} & =\int_{0}^{1} \frac{x}{(a x+1)\left(x^{2}+1\right)} d x \\ & =\frac{1}{a^{2}+1} \int_{0}^{1}\left(\frac{a}{x^{2}+1}+\frac{x}{x^{2}+1}-\frac{a}{a x+1}\right) d x \\ & =\frac{\pi a+2 \ln 2-4 \ln (a+1)}{4\left(a^{2}+1\right)}\end{aligned}由于 $I(0) = 0$，\begin{aligned} I(1) & =\int_{0}^{1}\left(\frac{\pi a+2 \ln 2-4 \ln (a+1)}{4\left(a^{2}+1\right)}\right) d a \\ & =\int_{0}^{1} \frac{\pi a+2 \ln 2}{4\left(a^{2}+1\right)} d a-\int_{0}^{1} \frac{\ln (a+1)}{\left(a^{2}+1\right)} d a \\ & =\int_{0}^{1} \frac{\pi a+2 \ln 2}{4\left(a^{2}+1\right)} d a-I(1) \\ \therefore I(1) & =\frac{1}{2} \int_{0}^{1} \frac{\pi a+2 \ln 2}{4\left(a^{2}+1\right)} d a\end{aligned}这是一个有理函数的积分，可以立即计算出来。结果是：$I(1)={\pi\ln2\over8}$. |
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青青子衿
Posted at 2023-4-12 10:48:56
isee
Posted at 2023-4-12 13:27:30
kuing
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