The integral we wish to evaluate is I = ∫ − ∞ ∞ sin m x x ( x 8 + a 8 ) d x I = \displaystyle \int_{-\infty}^{\infty} \dfrac{\sin{mx}}{x(x^8+a^8)} dx I = ∫ − ∞ ∞ x ( x 8 + a 8 ) sin m x d x a , m > 0 a, m \gt 0 a , m > 0
The best curve to use to evaluate this integral is likely this one:
We will use this contour integral: ∮ C e i m z z ( z 8 + a 8 ) d z \displaystyle \oint_C \dfrac{e^{imz}}{z(z^8+a^8)} dz ∮ C z ( z 8 + a 8 ) e i m z d z
The integrand of which has poles where:
z 8 + a 8 = 0 z 8 = − a 8 z 8 = a 8 ⋅ ( − 1 ) z n 8 = a 8 e π i ( 1 + 2 n ) , n ∈ Z z n = a e π i 8 ( 1 + 2 n ) . \begin{aligned} z^8 + a^8 &= 0 \\ z^8 &= -a^8 \\ z^8 &= a^8 \cdot (-1) \\ z^8_n &= a^8 e^{\pi i (1+2n)},\hspace{0.1cm} n\in \mathbb{Z} \\ z_n &= ae^{\dfrac{\pi i}{8} (1+2n)}. \end{aligned} z 8 + a 8 z 8 z 8 z n 8 z n = 0 = − a 8 = a 8 ⋅ ( − 1 ) = a 8 e π i ( 1 + 2 n ) , n ∈ Z = a e 8 π i ( 1 + 2 n ) . n = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 n=0,1,2,3,4,5,6,7 n = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 n = 0 , 1 , 2 , 3 n=0,1,2,3 n = 0 , 1 , 2 , 3 r n r_n r n z n z_n z n
r n = lim z → z n e i m z ( z − z 0 ) z ( z 8 + a 8 ) = lim z → z n e i m z z ⋅ 8 z 7 = e i m z n 8 z n 8 = − e i m z n 8 a 8 . \begin{aligned} r_n &= \lim_{z\rightarrow z_n} \dfrac{e^{imz} (z-z_0)}{z(z^8+a^8)} \\ &= \lim_{z\rightarrow z_n} \dfrac{e^{imz}}{z\cdot 8z^7} \\ &= \dfrac{e^{imz_n}}{8z^8_n} \\ &= -\dfrac{e^{imz_n}}{8a^8}. \end{aligned} r n = z → z n lim z ( z 8 + a 8 ) e i m z ( z − z 0 ) = z → z n lim z ⋅ 8 z 7 e i m z = 8 z n 8 e i m z n = − 8 a 8 e i m z n . Let us do some working to simplify r 0 = − e i m z 0 8 a 8 . r_0 = -\dfrac{e^{imz_0}}{8a^8}. r 0 = − 8 a 8 e i m z 0 . i m z 0 = a m e 5 π i 8 = − a m sin π 8 + a m i cos π 8 imz_0 = ame^{\dfrac{5\pi i}{8}} = -am\sin{\dfrac{\pi}{8}}+ami\cos{\dfrac{\pi}{8}} i m z 0 = a m e 8 5 π i = − a m sin 8 π + a m i cos 8 π p = sin π 8 p=\sin{\dfrac{\pi}{8}} p = sin 8 π q = cos π 8 q=\cos{\dfrac{\pi}{8}} q = cos 8 π
r 0 = − e − a m p + a m i q 8 a 8 = − e − a m p e a m i q 8 a 8 . r 1 = − e i m z 1 8 a 8 . \begin{aligned} r_0 &= -\dfrac{e^{-amp+amiq}}{8a^8} \\ &= -\dfrac{e^{-amp}e^{amiq}}{8a^8}.\\ r_1 &= -\dfrac{e^{imz_1}}{8a^8}. \end{aligned} r 0 r 1 = − 8 a 8 e − a m p + a m i q = − 8 a 8 e − a m p e a m i q . = − 8 a 8 e i m z 1 . Let us do some working to simplify this last line. i m z 1 = a m e 7 π i 8 = − a m cos π 8 + a m i sin π 8 = − a m q + a m i p imz_1 = ame^{\dfrac{7\pi i}{8}} = -am\cos{\dfrac{\pi}{8}}+ami\sin{\dfrac{\pi}{8}} = -amq+amip i m z 1 = a m e 8 7 π i = − a m cos 8 π + a m i sin 8 π = − a m q + a m i p
r 1 = − e − a m q + a m i p 8 a 8 = − e − a m q e a m i p 8 a 8 . r 2 = − e i m z 2 8 a 8 . \begin{aligned} r_1 &= -\dfrac{e^{-amq+amip}}{8a^8} \\ &= -\dfrac{e^{-amq}e^{amip}}{8a^8}.\\ r_2 &= -\dfrac{e^{imz_2}}{8a^8}. \end{aligned} r 1 r 2 = − 8 a 8 e − a m q + a m i p = − 8 a 8 e − a m q e a m i p . = − 8 a 8 e i m z 2 . Working on this last line. i m z 2 = a m e 9 π i 8 = − a m q − a m i p imz_2 = ame^{\dfrac{9\pi i}{8}} = -amq-amip i m z 2 = a m e 8 9 π i = − a m q − a m i p
r 2 = − e − a m q − a m i p 8 a 8 = − e − a m q e − a m i p 8 a 8 . r 3 = − e i m z 3 8 a 8 . \begin{aligned} r_2 &= -\dfrac{e^{-amq-amip}}{8a^8} \\ &= -\dfrac{e^{-amq}e^{-amip}}{8a^8}.\\ r_3 &= -\dfrac{e^{imz_3}}{8a^8}. \end{aligned} r 2 r 3 = − 8 a 8 e − a m q − a m i p = − 8 a 8 e − a m q e − a m i p . = − 8 a 8 e i m z 3 . Working on this last line. i m z 3 = a m e 11 π i 8 = − a m p − a m i q imz_3 = ame^{\dfrac{11\pi i}{8}} = -amp-amiq i m z 3 = a m e 8 1 1 π i = − a m p − a m i q
r 3 = − e − a m p − a m i q 8 a 8 = − e − a m p e − a m i q 8 a 8 . \begin{aligned} r_3 &= -\dfrac{e^{-amp-amiq}}{8a^8} \\ &= -\dfrac{e^{-amp}e^{-amiq}}{8a^8}. \end{aligned} r 3 = − 8 a 8 e − a m p − a m i q = − 8 a 8 e − a m p e − a m i q . Thus the sum of the residues of the poles within the region enclosed by C is:
∑ n = 0 3 r n = − 1 8 a 8 ( e − a m p e a m i q + e − a m q e a m i p + e − a m q e − a m i p + e − a m p e − a m i q ) = − 1 8 a 8 ( e − a m p ( e a m i q + e − a m i q ) + e − a m q ( e a m i p + e − a m i p ) ) = − 1 8 a 8 ( 2 e − a m p cos a m q + 2 e − a m q cos a m p ) = − 1 4 a 8 ( e − a m p cos a m q + e − a m q cos a m p ) . \begin{aligned} \sum_{n=0}^3 r_n &= -\dfrac{1}{8a^8} \left(e^{-amp}e^{amiq} + e^{-amq}e^{amip} + e^{-amq}e^{-amip} + e^{-amp}e^{-amiq}\right) \\ &= -\dfrac{1}{8a^8} \left(e^{-amp}\left(e^{amiq}+e^{-amiq}\right)+e^{-amq}\left(e^{amip}+e^{-amip}\right)\right) \\ &= -\dfrac{1}{8a^8} \left(2e^{-amp}\cos{amq} + 2e^{-amq}\cos{amp}\right) \\ &= -\dfrac{1}{4a^8} \left(e^{-amp}\cos{amq} + e^{-amq}\cos{amp}\right). \end{aligned} n = 0 ∑ 3 r n = − 8 a 8 1 ( e − a m p e a m i q + e − a m q e a m i p + e − a m q e − a m i p + e − a m p e − a m i q ) = − 8 a 8 1 ( e − a m p ( e a m i q + e − a m i q ) + e − a m q ( e a m i p + e − a m i p ) ) = − 8 a 8 1 ( 2 e − a m p cos a m q + 2 e − a m q cos a m p ) = − 4 a 8 1 ( e − a m p cos a m q + e − a m q cos a m p ) . Therefore our contour integral is:
∮ C e i m z z ( z 8 + a 8 ) d z = 2 π i ⋅ − 1 4 a 8 ( e − a m p cos a m q + e − a m q cos a m p ) = − π i 2 a 8 ( e − a m p cos a m q + e − a m q cos a m p ) . \begin{aligned} \oint_C \dfrac{e^{imz}}{z(z^8+a^8)} dz &= 2\pi i \cdot -\dfrac{1}{4a^8} \left(e^{-amp}\cos{amq} + e^{-amq}\cos{amp}\right) \\ &= -\dfrac{\pi i}{2a^8} \left(e^{-amp}\cos{amq} + e^{-amq}\cos{amp}\right). \end{aligned} ∮ C z ( z 8 + a 8 ) e i m z d z = 2 π i ⋅ − 4 a 8 1 ( e − a m p cos a m q + e − a m q cos a m p ) = − 2 a 8 π i ( e − a m p cos a m q + e − a m q cos a m p ) . Splitting our contour integral into subintegrals, we obtain:
∮ C e i m z z ( z 8 + a 8 ) d z = ∫ A B + ∫ B D + ∫ D E + ∫ E A \begin{aligned} \oint_C \dfrac{e^{imz}}{z(z^8+a^8)} dz &= \int_{AB} + \int_{BD} + \int_{DE} + \int_{EA} \end{aligned} ∮ C z ( z 8 + a 8 ) e i m z d z = ∫ A B + ∫ B D + ∫ D E + ∫ E A where each integral on the right-hand side of course shares the same integrand as that on the left.
Along AB, z = x z=x z = x d z = d x dz=dx d z = d x x x x − R -R − R − ϵ -\epsilon − ϵ
∫ A B = ∫ − R − ϵ e i m x x ( x 8 + a 8 ) d x . \begin{aligned} \int_{AB} = \int_{-R}^{-\epsilon} \dfrac{e^{imx}}{x(x^8+a^8)} dx. \end{aligned} ∫ A B = ∫ − R − ϵ x ( x 8 + a 8 ) e i m x d x . When ϵ → 0 \epsilon \rightarrow 0 ϵ → 0 R → ∞ R\rightarrow \infty R → ∞
∫ A B = ∫ − ∞ 0 e i m x x ( x 8 + a 8 ) d x . \int_{AB} = \int_{-\infty}^0 \dfrac{e^{imx}}{x(x^8+a^8)} dx. ∫ A B = ∫ − ∞ 0 x ( x 8 + a 8 ) e i m x d x . Along BD, z = ϵ e i θ z=\epsilon e^{i\theta} z = ϵ e i θ d z = i ϵ e i θ d θ dz=i\epsilon e^{i\theta} d\theta d z = i ϵ e i θ d θ θ \theta θ π \pi π 0 0 0
∫ B D = ∫ π 0 e i m ϵ e i θ ϵ e i θ ( ϵ 8 e 8 i θ + a 8 ) i ϵ e i θ d θ = − i ∫ 0 π e i m ϵ e i θ ϵ 8 e 8 i θ + a 8 d θ . \begin{aligned} \int_{BD} &= \int_{\pi}^0 \dfrac{e^{im\epsilon e^{i\theta}}}{\epsilon e^{i\theta}(\epsilon^8 e^{8i\theta}+a^8)} i\epsilon e^{i\theta} d\theta \\ &= -i \int_0^{\pi} \dfrac{e^{im\epsilon e^{i\theta}}}{\epsilon^8 e^{8i\theta}+a^8} d\theta. \end{aligned} ∫ B D = ∫ π 0 ϵ e i θ ( ϵ 8 e 8 i θ + a 8 ) e i m ϵ e i θ i ϵ e i θ d θ = − i ∫ 0 π ϵ 8 e 8 i θ + a 8 e i m ϵ e i θ d θ . When ϵ → 0 \epsilon \rightarrow 0 ϵ → 0
∫ B D = − i ∫ 0 π d θ a 8 = − π i a 8 . \begin{aligned} \int_{BD} &= -i \int_0^{\pi} \dfrac{d\theta}{a^8} \\ &= -\dfrac{\pi i}{a^8}. \end{aligned} ∫ B D = − i ∫ 0 π a 8 d θ = − a 8 π i . Along DE: z = x z=x z = x d z = d x dz=dx d z = d x x x x ϵ \epsilon ϵ R R R
∫ D E = ∫ ϵ R e i m x x ( x 8 + a 8 ) d x . \begin{aligned} \int_{DE} &= \int_{\epsilon}^R \dfrac{e^{imx}}{x(x^8+a^8)} dx. \end{aligned} ∫ D E = ∫ ϵ R x ( x 8 + a 8 ) e i m x d x . When ϵ → 0 \epsilon \rightarrow 0 ϵ → 0 R → ∞ R\rightarrow \infty R → ∞
∫ D E = ∫ 0 ∞ e i m x x ( x 8 + a 8 ) d x . \int_{DE} = \int_{0}^{\infty} \dfrac{e^{imx}}{x(x^8+a^8)} dx. ∫ D E = ∫ 0 ∞ x ( x 8 + a 8 ) e i m x d x . Finally, along EA, z = R e i θ z=Re^{i\theta} z = R e i θ d z = i R e i θ dz=iRe^{i\theta} d z = i R e i θ θ \theta θ 0 0 0 p i pi p i
∫ E A = ∫ 0 π e i m R e i θ R e i θ ( R 8 e 8 i θ + a 8 ) i R e i θ d θ = i ∫ 0 π e i m R cos θ − m R sin θ R 8 e 8 i θ + a 8 d θ = i ∫ 0 π e i m R cos θ e − m R sin θ R 8 e 8 i θ + a 8 d θ . \begin{aligned} \int_{EA} &= \int_{0}^{\pi} \dfrac{e^{imR e^{i\theta}}}{R e^{i\theta}(R^8 e^{8i\theta}+a^8)} iR e^{i\theta} d\theta \\ &= i \int_0^{\pi} \dfrac{e^{imR \cos{\theta}-mR\sin{\theta}}}{R^8 e^{8i\theta}+a^8} d\theta \\ &= i \int_0^{\pi} \dfrac{e^{imR \cos{\theta}}e^{-mR\sin{\theta}}}{R^8 e^{8i\theta}+a^8} d\theta. \end{aligned} ∫ E A = ∫ 0 π R e i θ ( R 8 e 8 i θ + a 8 ) e i m R e i θ i R e i θ d θ = i ∫ 0 π R 8 e 8 i θ + a 8 e i m R c o s θ − m R s i n θ d θ = i ∫ 0 π R 8 e 8 i θ + a 8 e i m R c o s θ e − m R s i n θ d θ . For 0 ≤ θ ≤ π 0\leq \theta \leq \pi 0 ≤ θ ≤ π e − m R sin θ ≤ 1 e^{-mR\sin{\theta}} \leq 1 e − m R s i n θ ≤ 1 ∣ e i m R cos θ ∣ = 1 |e^{imR\cos{\theta}}| = 1 ∣ e i m R c o s θ ∣ = 1 m ≥ 0 m \geq 0 m ≥ 0
∫ E A ≤ i ∫ 0 π d θ R 8 e 8 i θ + a 8 \begin{aligned} \int_{EA} \leq i \int_0^{\pi} \dfrac{d\theta}{R^8 e^{8i\theta}+a^8} \end{aligned} ∫ E A ≤ i ∫ 0 π R 8 e 8 i θ + a 8 d θ The magnitude of that denominator is:
∣ R 8 e 8 i θ + a 8 ∣ = ( R 8 cos 8 θ + a 8 ) 2 + R 16 sin 2 8 θ = R 16 + a 16 + 2 R 8 a 8 cos 8 θ ≥ R 16 + a 16 − 2 R 8 a 8 = ∣ R 8 − a 8 ∣ ∴ ∫ E A ≤ i ∫ 0 π d θ ∣ R 8 − a 8 ∣ = π i ∣ R 8 − a 8 ∣ . \begin{aligned} |R^8e^{8i\theta} + a^8| &= \sqrt{(R^8 \cos{8\theta}+a^8)^2 + R^{16} \sin^2{8\theta}} \\ &= \sqrt{R^{16} + a^{16} + 2R^8a^8 \cos{8\theta}} \\ &\geq \sqrt{R^{16}+a^{16} - 2R^8a^8} \\ &= |R^8-a^8| \\ \therefore \int_{EA} &\leq i\int_0^{\pi} \dfrac{d\theta}{|R^8-a^8|} \\ &= \dfrac{\pi i}{|R^8-a^8|}. \end{aligned} ∣ R 8 e 8 i θ + a 8 ∣ ∴ ∫ E A = ( R 8 cos 8 θ + a 8 ) 2 + R 1 6 sin 2 8 θ = R 1 6 + a 1 6 + 2 R 8 a 8 cos 8 θ ≥ R 1 6 + a 1 6 − 2 R 8 a 8 = ∣ R 8 − a 8 ∣ ≤ i ∫ 0 π ∣ R 8 − a 8 ∣ d θ = ∣ R 8 − a 8 ∣ π i . As R → ∞ R\rightarrow \infty R → ∞ ∫ E A → 0 \displaystyle \int_{EA} \rightarrow 0 ∫ E A → 0
Therefore the sum of our subintegrals (which is equal to our contour integral) is:
∮ C e i m z z ( z 8 + a 8 ) d z = ∫ − ∞ ∞ e i m x x ( x 8 + a 8 ) d x − π i a 8 \begin{aligned} \oint_C \dfrac{e^{imz}}{z(z^8+a^8)} dz &= \int_{-\infty}^{\infty} \dfrac{e^{imx}}{x(x^8+a^8)} dx -\dfrac{\pi i}{a^8} \end{aligned} ∮ C z ( z 8 + a 8 ) e i m z d z = ∫ − ∞ ∞ x ( x 8 + a 8 ) e i m x d x − a 8 π i Replacing the left-hand side with what the residue theorem tells us this contour integral equals yields:
− π i 2 a 8 ( e − a m p cos a m q + e − a m q cos a m p ) = ∫ − ∞ ∞ e i m x x ( x 8 + a 8 ) d x − π i a 8 ∴ ∫ − ∞ ∞ e i m x x ( x 8 + a 8 ) d x = π i 2 a 8 ( 2 − e − a m p cos a m q − e − a m q cos a m p ) ∫ − ∞ ∞ cos m x + i sin m x x ( x 8 + a 8 ) d x = π i 2 a 8 ( 2 − e − a m p cos a m q − e − a m q cos a m p ) ∫ − ∞ ∞ cos m x x ( x 8 + a 8 ) d x + i ∫ − ∞ ∞ sin m x x ( x 8 + a 8 ) d x = π i 2 a 8 ( 2 − e − a m p cos a m q − e − a m q cos a m p ) . \begin{aligned} -\dfrac{\pi i}{2a^8} \left(e^{-amp}\cos{amq} + e^{-amq}\cos{amp}\right) &= \int_{-\infty}^{\infty} \dfrac{e^{imx}}{x(x^8+a^8)} dx -\dfrac{\pi i}{a^8} \\ \therefore \int_{-\infty}^{\infty} \dfrac{e^{imx}}{x(x^8+a^8)} dx &= \dfrac{\pi i}{2a^8} \left(2 - e^{-amp}\cos{amq} - e^{-amq}\cos{amp}\right) \\ \int_{-\infty}^{\infty} \dfrac{\cos{mx}+i\sin{mx}}{x(x^8+a^8)} dx &= \dfrac{\pi i}{2a^8} \left(2 - e^{-amp}\cos{amq} - e^{-amq}\cos{amp}\right) \\ \int_{-\infty}^{\infty} \dfrac{\cos{mx}}{x(x^8+a^8)} dx + i\int_{-\infty}^{\infty} \dfrac{\sin{mx}}{x(x^8+a^8)} dx &= \dfrac{\pi i}{2a^8} \left(2 - e^{-amp}\cos{amq} - e^{-amq}\cos{amp}\right). \end{aligned} − 2 a 8 π i ( e − a m p cos a m q + e − a m q cos a m p ) ∴ ∫ − ∞ ∞ x ( x 8 + a 8 ) e i m x d x ∫ − ∞ ∞ x ( x 8 + a 8 ) cos m x + i sin m x d x ∫ − ∞ ∞ x ( x 8 + a 8 ) cos m x d x + i ∫ − ∞ ∞ x ( x 8 + a 8 ) sin m x d x = ∫ − ∞ ∞ x ( x 8 + a 8 ) e i m x d x − a 8 π i = 2 a 8 π i ( 2 − e − a m p cos a m q − e − a m q cos a m p ) = 2 a 8 π i ( 2 − e − a m p cos a m q − e − a m q cos a m p ) = 2 a 8 π i ( 2 − e − a m p cos a m q − e − a m q cos a m p ) . Equating the imaginary part of this expression yields:
∫ − ∞ ∞ sin m x x ( x 8 + a 8 ) d x = π 2 a 8 ( 2 − e − a m p cos a m q − e − a m q cos a m p ) = π 2 a 8 ( 2 − e − a m sin π 8 cos ( a m cos π 8 ) − e − a m cos π 8 cos ( a m sin π 8 ) ) . \begin{aligned} \int_{-\infty}^{\infty} \dfrac{\sin{mx}}{x(x^8+a^8)} dx &= \dfrac{\pi}{2a^8} \left(2 - e^{-amp}\cos{amq} - e^{-amq}\cos{amp}\right) \\ &= \dfrac{\pi}{2a^8} \left(2 - e^{-am\sin{\dfrac{\pi}{8}}}\cos{\left(am\cos{\dfrac{\pi}{8}}\right)} - e^{-am\cos{\dfrac{\pi}{8}}}\cos{\left(am\sin{\dfrac{\pi}{8}}\right)}\right). \end{aligned} ∫ − ∞ ∞ x ( x 8 + a 8 ) sin m x d x = 2 a 8 π ( 2 − e − a m p cos a m q − e − a m q cos a m p ) = 2 a 8 π ⎝ ⎜ ⎛ 2 − e − a m s i n 8 π cos ( a m cos 8 π ) − e − a m c o s 8 π cos ( a m sin 8 π ) ⎠ ⎟ ⎞ . Which is what we were required to determine.
