定义 \(f(t)\) 的拉普拉斯变换
\[\color{orchid}{\mathscr L[f(t)]=\int_0^\infty f(t)\text e^{-st}\text dt}\\
\]
记作
\[\color{darkorange}{F(s)=\mathscr L[f(t)]}\\
\]
并规定 \(t<0\) 时 \(f(t)=0\)
我们来近距离感受它。
\(\rm Qn1.\quad\)计算 \(\mathscr L[f'(t)]\)
\(\rm Sol\quad\) 采用分部积分 (IBP)
\[\begin{aligned}
\mathscr L[f'(t)]=&\int_0^\infty f'(t) \text e^{-st} \text dt\\
=& \int_0^\infty \text e^{-st} \text df(t)\\
=& f(t)\text e^{-st}\bigg|_0^\infty + s\int_0^\infty f(t) \text e^{-st} \text dt\\
=&-f(0)+sF(s)
\end{aligned}
\]
\(\rm Qn2.\quad\)计算 \(\mathscr L\left[\displaystyle{ \int_0^t f(t) dt }\right]\)
\(\rm Sol\quad\) 同样是 IBP
\[\begin{aligned}
\mathscr L\left[\int_0^t f(t)\text dt\right]=&-\frac1s\int_0^\infty\left(\int_0^t f(t)\text dt\right) \text d\text e^{-st}\\
=& -\frac1s\text e^{-st}\int_0^t f(t) \text dt\bigg|_0^\infty+\frac1s \int_0^\infty f(t) \text e^{-st} \text dt\\
=& \frac{F(s)}{s}
\end{aligned}
\]
\(\rm Qn3.\quad\)计算 \(\mathscr L\left[\dfrac{f(t)}{t}\right]\)
\(\rm Sol\quad\) 我们尝试对 \(s\) 求导后积分 (Feynman's Trick)
\[\begin{aligned}
\mathscr L\left[\dfrac{f(t)}{t}\right]=&\int_0^\infty \dfrac{f(t)}{t}\text e^{-st}\text dt\\=& \int_s^\infty \left(\int_0^\infty f(t)\text e^{-st}\text dt\right)\text ds\\
=& \int_s^\infty F(s) \text ds
\end{aligned}
\]
\(\rm Qn4.\quad\)计算 \(\mathscr L[\sin t]\)
\(\rm Sol\quad\) 可通过欧拉公式简化运算
\[\begin{aligned}
\mathscr L[\sin t]=&\int_0^\infty \sin t \text e^{-st} \text dt\\
=&\text{Im} \int_0^\infty \text e^{(\text i-s)t} \text dt\\
=&\text{Im} \frac{1}{s-\text i}\\
=&\frac{1}{s^2+1}
\end{aligned}
\]
到这里我们已经足够处理一些问题了。
\(\rm Qn5.\quad\)计算 \(\displaystyle{ \int_0^\infty \frac{\sin t}{t} \text dt }\)
\(\rm Sol\quad\) 加入参数 \(\text e^{-st}\) 使之符合拉普拉斯变换的形式
\[\begin{aligned}
\int_0^\infty \frac{\sin t}{t}\text e^{-st}\text dt=&\mathscr L\left[\frac{\sin t}{t}\right]\\
=& \int_s^\infty \mathscr L[\sin t]\text ds\\
=& \int_s^\infty \frac{1}{s^2+1}\text ds\\
=& \frac{\pi}{2}-\arctan(s)
\end{aligned}
\]
代入 \(s=0\) 得到
\[\int_0^\infty \frac{\sin t}{t}\text dt=\frac{\pi}{2}
\]
\(\rm Qn6.\quad\)计算 \(\displaystyle{ \int_0^\infty \frac{|\cos t|}{\text e^t} \text dt }\)
\(\rm Sol\quad\) 我们发现 \(f(t)=|\cos t|\) 是周期函数
\[\begin{aligned}
\int_0^\infty f(t)\text e^{-st}\text dt=&\int_0^T f(t)\text e^{-st} \text dt+\int_T^\infty f(t)\text e^{-st} \text dt\\
=& \int_0^T f(t)\text e^{-st} \text dt+\int_0^\infty f(t)\text e^{-s(t+T)} \text dt\\
=&\int_0^T f(t)\text e^{-st} \text dt+\text e^{-sT}\int_0^\infty f(t)\text e^{-st} \text dt
\end{aligned}
\]
即
\[\int_0^\infty f(t)\text e^{-st}\text dt=\frac{1}{1-\text e^{-sT}}\int_0^T f(t)\text e^{-st} \text dt
\]
那么
\[\begin{aligned}
\int_0^\infty \frac{|\cos t|}{\text e^t} dt=&\frac{1}{1-\text e^{-\pi}}\int_0^{\pi} \frac{|\cos t|}{\text e^t}\text dt\\
=&\frac{1}{1-\text e^{-\pi}}\text{Re} \left(\int_0^\frac{\pi}2 \text e^{(\text i-1)t}\text dt-\int_\frac{\pi}2^\pi \text e^{(\text i-1)t}\text dt\right)\\=&\frac12+\frac12\text{csch}\frac{\pi}2
\end{aligned}
\]
\(\rm Qn7.\quad\)计算 \(\displaystyle{ \int_0^\infty\dfrac{\cos bt}{a^2 +t^2}}\text dt\)
\(\rm Sol\quad\) 采用含参积分的策略，将 \(b\) 视作变量进行拉普拉斯变换
\[\begin{aligned}
\mathscr L\left[\int_0^\infty \frac{\cos bt}{a^2 +t^2}\text dt\right]=&\int_0^\infty \frac{1}{a^2+x^2}\left(\int_0^\infty \cos bt \text e^{-sb}\text db\right)\text dt\\
=&\int_0^\infty \frac{s}{(a^2+t^2)(s^2+t^2)}\text dt\\
=&\frac{\pi}{2a(s+a)}
\end{aligned}
\]
我们得到了一个简洁的表达式。看起来需要对 \(\dfrac{\pi}{2a(s+a)}\) 进行逆变换。
我们对拉普拉斯逆变换作出定义
\[\color{orchid}{\mathscr L^{-1}[F(s)]=\frac{1}{2\pi\text i}\int_{\beta-\text i\infty}^{\beta+\text i\infty}F(s)\text e^{st}\text ds}
\]
记作
\[\color{darkorange}{f(t)=\mathscr L^{-1}[F(s)]}
\]
采取围道积分 (Bromwich Contour) 可以证明
\[\mathscr L^{-1}[F(s)] = \sum_{k=1}^n \text{Res} [F(s)\text e^{st},s_k]
\]
亦即逆变换的结果为所有孤立奇点留数之和。
因此
\[\begin{aligned}
\mathscr L^{-1}\left[\frac{\pi}{2a(s+a)}\right]=&\text{Res}\left[\frac{\pi\text e^{sb}}{2a(s+a)},-a\right]\\
=&\frac{\pi}{2a}\text e^{-ab}
\end{aligned}
\]
没有学习复变函数也不用担心，下面是常用拉普拉斯变换表
\[\begin{aligned} &\mathscr L[1]=\frac{1}{s}\\ &\mathscr L[t]=\frac{1}{s^2}\\ &\mathscr L[t^n]=\frac{n!}{s^{n+1}}\\ &\mathscr L[\text e^{-\alpha t}]=\frac{1}{s+\alpha}\\ &\mathscr L[t\text e^{-\alpha t}]=\frac{1}{(s+\alpha)^2}\\ &\mathscr L[t^n\text e^{-\alpha t}]=\frac{n!}{(s+\alpha)^{n+1}} \\ &\mathscr L[\sin\alpha t]=\frac{\alpha}{s^2+\alpha^2}\\ &\mathscr L[\cos\alpha t]=\frac{s}{s^2+\alpha^2} \end{aligned}
\]
我们直接得到
\[\frac{\pi}{2a(s+a)}=\frac{\pi}{2a}\mathscr L[\text e^{-ab}]
\]
更复杂的形式可以用部分分式展开。
\(\rm Qn8.\quad\)计算 \(\displaystyle{\int_0^\infty \frac{t-\sin t}{t^3(1+t^2)}\text dt}\)
\(\rm Sol\quad\) 我们采取交换乘积的技巧 (展开即证)
\[\int_0^\infty F(x)g(x)\text dx=\int_0^\infty f(x)G(x)\text dx
\]
由于
\[\mathscr L[t-\sin t]=\frac{1}{x^2(1+x^2)},\ \ \mathscr L^{-1}\left[\frac{1}{t^3(1+t^2)}\right]=\frac{x^2}2+\cos x-1
\]
所以
\[\begin{aligned}
\int_0^\infty \frac{t-\sin t}{t^3(1+t^2)}\text dt=&\int_0^\infty \frac{1}{x^2(1+x^2)}(\frac{x^2}2+\cos x-1)\text dx\\
=&\frac{3\pi}{4}-\int_0^\infty\frac{\cos x}{x^2+1}\text dx-\int_0^\infty \frac{\sin x}x \text dx\\
=&\frac{\pi}4-\frac{\pi}{2\text e}
\end{aligned}
\]
\(\rm Qn9.\quad\)证明 \(\text{B}(p,q)=\dfrac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\)
\(\rm Sol\quad\) 定义欧拉积分
\[\begin{aligned}
&\Gamma(s)=\int_0^\infty t^{s-1}\text e^{-t}\text dt\\
&\text{B}(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1} \text dt
\end{aligned}
\]
其中 \(s\) 为整数时 \(\Gamma(s)=(s-1)!\)
我们需要证明
\[\int_0^1 t^{p-1}(1-t)^{q-1} \text dt=\dfrac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}
\]
由变换表
\[\mathscr L[t^{n-1}]=\frac{\Gamma(n)}{s^n}
\]
右边可以写成
\[\frac{\mathscr L[t^{p-1}]\mathscr L[t^{q-1}]}{\mathscr L[t^{p+q-1}]}
\]
我们尝试计算分子
\[\begin{aligned}
\mathscr L[t^{p-1}]\mathscr L[t^{q-1}]=&\int_0^\infty\tau^{p-1}\text e^{-s\tau}\text d\tau\int_0^\infty t^{q-1}\text e^{-st}\text dt\\
=&\int_0^\infty\int_0^\infty \tau^{p-1}t^{q-1}\text e^{-s(\tau+t)}\text d\tau\text dt\\
=&\int_0^\infty\int_0^\infty \tau^{p-1}(t-\tau)^{q-1}\text e^{-st}\text d\tau\text dt\\
=&\int_0^\infty \left(\int_0^\infty \tau^{p-1}(t-\tau)^{q-1}\text d\tau\right)\text e^{-st}\text dt
\end{aligned}
\]
规定 \(t<0\) 时 \(t^{p-1} =t^{q-1}=0\)，因此改写为
\[\mathscr L\left[\int_0^t \tau^{p-1}(t-\tau)^{q-1}\text d\tau\right]
\]
这已经是原式左边的形式了呀。只需要进行一次逆变换就可以了。
\[\begin{aligned}
\int_0^t \tau^{p-1}(t-\tau)^{q-1} \text d\tau=&\mathscr L^{-1}[\mathscr L[t^{p-1}]\mathscr L[t^{q-1}]]\\=&\mathscr L^{-1}\left[\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\mathscr L[t^{p+q-1}]\right]\\
=&\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}t^{p+q-1}
\end{aligned}
\]
代入 \(t=1\) 得到
\[\text{B}(p,q)=\dfrac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}
\]
我们实质上证明了卷积定理 (Convolution Theorem)
\[\mathscr L[f(t)*g(t)]=\mathscr L[f(t)]\mathscr L[g(t)]
\]
其中 \(*\) 指卷积运算
\[f(t)*g(t)=\int_{-\infty}^\infty f(\tau)g(t-\tau)\text d\tau
\]
\(\rm Qn10.\quad\)计算 \(\displaystyle{\int_0^\infty\cos(t^k)\text dt}\)
\(\rm Sol\quad\)
\[\int_0^\infty\cos(t^k)\text dt=\frac1k\int_0^\infty \frac{\cos t}{t^{1-\frac1k}}\text dt
\]
由于
\[\mathscr L[\cos t]=\frac{x}{x^2+1},\ \ \mathscr L^{-1}\left[\frac{1}{t^{1-\frac1k}}\right]=\frac{x^{-\frac1k}}{\Gamma(1-\frac1k)}
\]
交换顺序并换元 \(\dfrac{x^2}{1+ x^2} \mapsto x\) 得到
\[\frac{1}{k\Gamma(1-\frac1k)}\int_0^\infty \frac{x^{1-\frac1k}}{x^2+1}\text dx=\frac{1}{2k\Gamma(1-\frac1k)}\int_0^1 x^{-\frac1{2k}}(1-x)^{\frac1{2k}-1}\text dx
\]
利用上一题的结论和欧拉反射公式 (Euler's Reflection Formula)
\[\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}
\]
化简得
\[\begin{aligned}
\frac{1}{2k\Gamma(1-\frac1k)}\Gamma(\frac{1}{2k})\Gamma(1-\frac1{2k})=&\frac{1}{2k\Gamma(1-\frac1k)}\frac{\pi}{\sin(1-\frac1{2k})\pi}\\=&\frac1k\Gamma(\frac1k)\cos\frac{\pi}{2k}
\end{aligned}
\]
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