积分练习 2
积分练习 2
1
求积分
$$
I=\int_0^1\dfrac{x\ln(1+x)}{(1+x^2)^2}\text{d}x\
$$
【Sol】:
$\begin{aligned}
I=&\int_0^1\ln(1+x)\text{d}\left[-\dfrac{1}{2(1+x^2)} \right]\\
=&-\dfrac{\ln(1+x)}{2(1+x^2)}\bigg|_0^1+\dfrac{1}{2}\int_0^1\dfrac{1}{(1+x^2)(1+x)}\text{d}x\\
=&-\dfrac{1}{4}\ln2+\dfrac{1}{4}\int_0^1\left[\dfrac{1}{1+x}+\dfrac{1-x}{1+x^2} \right]\\
=&-\dfrac{1}{4}\ln2+\dfrac{1}{4}\left[\ln(1+x)+\arctan x-\dfrac{1}{2}\ln(1+x^2) \right]\bigg|_0^1\\
=&-\dfrac{1}{8}\ln2+\dfrac{\pi}{16}
\end{aligned}$
2
求积分 $$ I=\int_0^{\infty} \dfrac{e^{-x}-e^{-2x}}{x}\sin x\text{d}x $$
【Sol】: 用拉普拉斯变换求
令 $f(x)=(e^{-x}-e^{-2x})\sin x$
$F(x)=\dfrac{1}{(x+1)^2+1}-\dfrac{1}{(x+2)^2+1}$
$\begin{aligned}
I=&\int_0^{\infty}\left[\dfrac{1}{(x+1)^2+1}-\dfrac{1}{(x+2)^2+1} \right]\text{d}x\\
=&\left[\arctan(x+1)-\arctan(x+2) \right]\bigg|_{0}^{\infty}\\
=&\arctan 2-\dfrac{\pi}{4}
\end{aligned}$
3
求积分
$$ I=\int_0^1\dfrac{\ln(x+\sqrt{1-x^2})}{x}\text{d}x $$
【Sol】: 考虑含参积分
$\displaystyle I(a)=\int_0^1\dfrac{\ln(xa+\sqrt{1-x^2})}{x}\text{d}x,;;I=I(1)$
$\begin{aligned}
I(0)=&\int_0^{1}\dfrac{\ln\sqrt{1-x^2}}{x}\text{d}x\\
=&\dfrac{1}{2}\left[\int_0^1\dfrac{\ln(1+x)}{x}\text{d}x+\int_0^{1}\dfrac{\ln(1-x)}{x}\text{d}x \right]\\
=&\dfrac{1}{2}\left[\dfrac{\pi^2}{12}-\dfrac{\pi^2}{6} \right]\\
=&-\dfrac{\pi^2}{24}
\end{aligned}$
$\begin{aligned}
I'(a)=&\int_0^1\dfrac{1}{xa+\sqrt{1-x^2}}\text{d}x\\
=&\int_0^{\frac{\pi}{2}}\dfrac{1}{a\sin t+\cos t}\cos t\text{d}t\\
=&\dfrac{1}{a^2+1}\left[ \int_0^{\frac{\pi}{2}}\dfrac{(a\sin t+\cos t)+a(a\cos t-\sin t)}{a\sin t+\cos t}\text{d}t\right]\\
=&\dfrac{1}{a^2+1}\left[\dfrac{\pi}{2}+a\ln a \right]
\end{aligned}$
$\begin{aligned}
I=&I(1)-I(0)\\
=&I(0)+\int_0^1I'(a)\text{d}a\\
=&-\dfrac{\pi^2}{24}+\int_0^{1}\dfrac{1}{a^2+1}\left[\dfrac{\pi}{2}+a\ln a \right]\text{d}a\\
=&-\dfrac{\pi^2}{24}+\dfrac{\pi^2}{8}+\int_0^1\dfrac{a\ln a}{a^2+1}\text{d}a\\
=&\dfrac{\pi^2}{12}+\dfrac{1}{2}\left[ \ln a\ln(a^2+1)\bigg|_0^1-\int_0^1\dfrac{\ln(a^2+1)}{a}\text{d}a\right]\\
=&\dfrac{\pi^2}{12}+\dfrac{1}{2}\left[-\dfrac{1}{2}\dfrac{\pi^2}{12} \right]\\
=&\dfrac{\pi^2}{16}
\end{aligned}$
来自 函数君爱复艾克斯