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九州大学 システム情報科学府 情報学専攻・情報知能工学専攻・電気電子工学専攻 共通 2010年度入学 数学 問4（複素関数論）

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Author​

思齐塾, 祭音Myyura

Description​

次の積分の値を複素積分によって求めよ。 ただし、 0<m<n0 < m < n である。

∫−∞∞emx1+enxdx\int_{-\infty}^{\infty} \frac{e^{mx}}{1 + e^{nx}} dx

题目描述​

设参数 m,nm,n 满足 0<m<n0<m<n。使用复积分方法求广义积分

∫−∞∞emx1+enx dx\int_{-\infty}^{\infty}\frac{e^{mx}}{1+e^{nx}}\,dx

的值。

Kai​

Let f(z)=emz1+enzf(z) = \frac{e^{mz}}{1+e^{nz}} . We consider the rectangular contour CRC_R with vertices at −R,R,R+i2πn,−R+i2πn-R, R, R + i\frac{2\pi}{n}, -R + i\frac{2\pi}{n} for a large R>0R>0 . The integral around this contour is given by

∮CRemz1+enzdz=∫−RRemx1+enxdx+∫02πnem(R+iy)1+en(R+iy)idy+∫R−Rem(x+i2πn)1+en(x+i2πn)dx+∫2πn0em(−R+iy)1+en(−R+iy)idy\oint_{C_R} \frac{e^{mz}}{1+e^{nz}}dz = \int_{-R}^{R} \frac{e^{mx}}{1+e^{nx}}dx + \int_{0}^{\frac{2\pi}{n}} \frac{e^{m(R+iy)}}{1+e^{n(R+iy)}}idy + \int_{R}^{-R} \frac{e^{m(x+i\frac{2\pi}{n})}}{1+e^{n(x+i\frac{2\pi}{n})}}dx + \int_{\frac{2\pi}{n}}^{0} \frac{e^{m(-R+iy)}}{1+e^{n(-R+iy)}}idy

As R→∞R \to \infty , the second and the fourth integrals vanish. The third integral can be rewritten as

∫R−Rem(x+i2πn)1+en(x+i2πn)dx=−∫−RRemxei2πmn1+enxei2πdx=−ei2πmn∫−RRemx1+enxdx\int_{R}^{-R} \frac{e^{m(x+i\frac{2\pi}{n})}}{1+e^{n(x+i\frac{2\pi}{n})}}dx = -\int_{-R}^{R} \frac{e^{mx} e^{i\frac{2\pi m}{n}}}{1+e^{nx} e^{i2\pi}}dx = -e^{i\frac{2\pi m}{n}} \int_{-R}^{R} \frac{e^{mx}}{1+e^{nx}}dx

Taking R→∞R\to\infty,

lim⁡R→∞∮CRemz1+enzdz=(1−ei2πmn)∫−∞∞emx1+enxdx\lim_{R\to\infty}\oint_{C_R} \frac{e^{mz}}{1+e^{nz}}dz = (1 - e^{i\frac{2\pi m}{n}}) \int_{-\infty}^{\infty} \frac{e^{mx}}{1+e^{nx}}dx

The pole of f(z)f(z) inside the contour is at z=iπnz = i\frac{\pi}{n} . The residue at this pole is given by

Res(f,iπn)=lim⁡z→iπn(z−iπn)emz1+enz=lim⁡z→iπnemznenz=eiπmnneiπ=−eiπmnn\text{Res}(f, i\frac{\pi}{n}) = \lim_{z \to i\frac{\pi}{n}} (z - i\frac{\pi}{n}) \frac{e^{mz}}{1+e^{nz}} = \lim_{z \to i\frac{\pi}{n}} \frac{e^{mz}}{ne^{nz}} = \frac{e^{i\frac{\pi m}{n}}}{ne^{i\pi}} = -\frac{e^{i\frac{\pi m}{n}}}{n}

By residue theorem,

∮CRemz1+enzdz=2πiRes(f,iπn)=−2πieiπmnn\oint_{C_R} \frac{e^{mz}}{1+e^{nz}}dz = 2\pi i \text{Res}(f, i\frac{\pi}{n}) = -2\pi i \frac{e^{i\frac{\pi m}{n}}}{n}

Thus,

(1−ei2πmn)∫−∞∞emx1+enxdx=−2πineiπmn(1 - e^{i\frac{2\pi m}{n}}) \int_{-\infty}^{\infty} \frac{e^{mx}}{1+e^{nx}}dx = -\frac{2\pi i}{n} e^{i\frac{\pi m}{n}}
∫−∞∞emx1+enxdx=−2πineiπmn1−ei2πmn=2πineiπmnei2πmn−1=2πin1eiπmn−e−iπmn=2πin12isin⁡(πmn)=πnsin⁡(πmn)\int_{-\infty}^{\infty} \frac{e^{mx}}{1+e^{nx}}dx = -\frac{2\pi i}{n} \frac{e^{i\frac{\pi m}{n}}}{1 - e^{i\frac{2\pi m}{n}}} = \frac{2\pi i}{n} \frac{e^{i\frac{\pi m}{n}}}{e^{i\frac{2\pi m}{n}} - 1} = \frac{2\pi i}{n} \frac{1}{e^{i\frac{\pi m}{n}} - e^{-i\frac{\pi m}{n}}} = \frac{2\pi i}{n} \frac{1}{2i \sin(\frac{\pi m}{n})} = \frac{\pi}{n \sin(\frac{\pi m}{n})}
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