6.1 參考解答

  1. (c) ∫0π2cosh⁡(it) dt=−isinh⁡it ∣0π2\int_{0}^{\frac{\pi}{2}} \cosh{(it)} \space dt=-i\sinh{it}\space|_{0}^{\frac{\pi}{2}}

    =−i[sinh⁡(iπ2)−sinh⁡(i⋅0)]=-i[\sinh{(i\frac{\pi}{2})}-\sinh{(i \cdot 0)}]

    =−i[sinh⁡0cos⁡π2+icosh⁡0sin⁡π2]=-i[\sinh{0}\cos{\frac{\pi}{2}}+i\cosh{0}\sin{\frac{\pi}{2}}]

    =−i⋅i=-i \cdot i

    =1=1

    (e) ∫0π4teit dt=∫0π4t(cos⁡t+isin⁡t) dt\int_{0}^{\frac{\pi}{4}} te^{it} \space dt=\int_{0}^{\frac{\pi}{4}} t(\cos{t}+i\sin{t}) \space dt

    =∫0π4tcos⁡t dt+i∫0π4sin⁡t dt=\int_{0}^{\frac{\pi}{4}} t\cos{t} \space dt+i\int_{0}^{\frac{\pi}{4}} \sin{t} \space dt

    =[tsin⁡t ∣0π4−∫0π4sin⁡t dt]+i[−tcos⁡t ∣0π4+∫0π4cos⁡t dt]=[t\sin{t}\space |_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \sin{t} \space dt]+i[-t\cos{t}\space |_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}} \cos{t} \space dt]

    =[(π4sin⁡π4−0)−(−cos⁡t ∣0π4)]+i[(−π4cos⁡π4−0)+(sin⁡t ∣0π4)]=[(\frac{\pi}{4}\sin{\frac{\pi}{4}}-0)-(-\cos{t}\space |_{0}^{\frac{\pi}{4}})]+i[(-\frac{\pi}{4}\cos{\frac{\pi}{4}}-0)+(\sin{t}\space |_{0}^{\frac{\pi}{4}})]

    =2π8−(−cos⁡π4+cos⁡0)+i[−2π8+(sin⁡π4−sin⁡0)]=\frac{\sqrt{2}\pi}{8}-(-\cos{\frac{\pi}{4}}+\cos{0})+i[-\frac{\sqrt{2}\pi}{8}+(\sin{\frac{\pi}{4}}-\sin{0})]

    =2π8+22−1+i(22−2π8)=\frac{\sqrt{2}\pi}{8}+\frac{\sqrt{2}}{2}-1+i(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}\pi}{8})

  1. 1。1^。 m=nm=n

    ∫02πeimt⋅e−int dt=∫02πe0 dt=t ∣02π=2π\int_{0}^{2\pi} e^{imt} \cdot e^{-int} \space dt=\int_{0}^{2\pi} e^0 \space dt=t \space |_{0}^{2\pi}=2\pi 

    2。2^。 m≠nm \neq n

    ∫02πeimt⋅e−int dt\int_{0}^{2\pi} e^{imt} \cdot e^{-int} \space dt

    =∫02πei(m−n)t dt=\int_{0}^{2\pi} e^{i(m-n)t} \space dt

    =∫02πcos⁡(m−n)t+isin⁡(m−n)t dt=\int_{0}^{2\pi} \cos{(m-n)t}+i\sin{(m-n)t} \space dt

    =∫02πcos⁡(m−n)t dt+i∫02πsin⁡(m−n)t dt=\int_{0}^{2\pi} \cos{(m-n)t} \space dt+i\int_{0}^{2\pi} \sin{(m-n)t} \space dt

    =1m−nsin⁡(m−n)t ∣02π−i1m−ncos⁡(m−n)t ∣02π=\frac{1}{m-n}\sin{(m-n)t} \space |_{0}^{2\pi}-i\frac{1}{m-n}\cos{(m-n)t} \space |_{0}^{2\pi}

    =1m−n[(sin⁡(m−n)2π−sin⁡(m−n)0)−i(cos⁡(m−n)2π−cos⁡(m−n)0)]=\frac{1}{m-n}[(\sin{(m-n)2\pi}-\sin{(m-n)0})-i(\cos{(m-n)2\pi}-\cos{(m-n)0})]

    =0=0

  1. ∫0∞e−zt dt=−1ze−zt ∣t=0t=∞\int_{0}^{\infty} e^{-zt}\space dt=-\frac{1}{z}e^{-zt}\space |_{t=0}^{t=\infty}

    =−1z(lim⁡t→∞e−zt−lim⁡t→0e−zt)=-\frac{1}{z}(\lim_{t \to \infty} e^{-zt}-\lim_{t \to 0} e^{-zt}) if Re(z)<0⇒lim⁡t→∞e−zt\text{Re}(z)<0 \Rightarrow \lim_{t \to \infty} e^{-zt} diverges

    =−1z(0−1)=-\frac{1}{z}(0-1)

    =1z=\frac{1}{z}

5. ∗∫abf(t) dt=F(t) ∣ab=F(b)−F(a)\ast \int_{a}^{b} f(t)\space dt=F(t) \space |_{a}^{b}=F(b)-F(a)

ddt[12(f(t))2]=f(t)f’(t)\frac{d}{dt}[\frac{1}{2}(f(t))^2]=f(t)f’(t)

∫abf(t)f’(t) dt=12(f(t))2 ∣ab=12(f(b))2−12(f(a))2\int_{a}^{b} f(t)f’(t)\space dt=\frac{1}{2}(f(t))^2\space |_{a}^{b}=\frac{1}{2}(f(b))^2-\frac{1}{2}(f(a))^2

  1. i∫0π2etsin⁡t dt=i[etsin⁡t ∣0π2−∫0π2etcos⁡t dt]i\int_{0}^{\frac{\pi}{2}} e^t\sin{t}\space dt=i[e^t\sin{t}\space |_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} e^t\cos{t}\space dt]

    =i[(eπ2sin⁡π2−e0sin⁡0)−∫0π2etcos⁡t dt]=i[(e^{\frac{\pi}{2}}\sin{\frac{\pi}{2}}-e^0\sin{0})-\int_{0}^{\frac{\pi}{2}} e^t\cos{t}\space dt]

    =i[eπ2−(etcos⁡t ∣0π2+∫0π2etsin⁡t dt)]=i[e^{\frac{\pi}{2}}-(e^t\cos{t}\space |_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}} e^t\sin{t}\space dt)]

    =i[eπ2−[(eπ2cos⁡π2−e0cos⁡0)+∫0π2etsin⁡t dt]]=i[e^{\frac{\pi}{2}}-[(e^{\frac{\pi}{2}}\cos{\frac{\pi}{2}}-e^0\cos{0})+\int_{0}^{\frac{\pi}{2}} e^t\sin{t}\space dt]]

    =i[(eπ2+1)−∫0π2etsin⁡t dt]=i[(e^{\frac{\pi}{2}}+1)-\int_{0}^{\frac{\pi}{2}} e^t\sin{t}\space dt]

    ⇒2i∫0π2etsin⁡t dt=i(eπ2+1)\Rightarrow 2i\int_{0}^{\frac{\pi}{2}} e^t\sin{t}\space dt=i(e^{\frac{\pi}{2}}+1)

    ⇒i∫0π2etsin⁡t dt=i2(eπ2+1)\Rightarrow i\int_{0}^{\frac{\pi}{2}} e^t\sin{t}\space dt=\frac{i}{2}(e^{\frac{\pi}{2}}+1)