In this article, we shall evaluate the following integral using contour integration methods.
I = ∫ − ∞ ∞ x 2 cos m x ( x 4 + a 4 ) 2 d x I = \int_{-\infty}^{\infty} \dfrac{x^2\cos{mx}}{(x^4+a^4)^2} dx I = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 cos m x d x Our first couple of questions, which are impossible to separate from one another as they are closely connected, is which curve and which integrand should we use to determine the value of I I I
The contour must have a component whose line integral evaluates to either I I I ∫ 0 ∞ x 2 cos m x ( x 4 + a 4 ) 2 d x = 1 2 I \displaystyle \int_{0}^{\infty} \dfrac{x^2\cos{mx}}{(x^4+a^4)^2} dx = \dfrac{1}{2} I ∫ 0 ∞ ( x 4 + a 4 ) 2 x 2 cos m x d x = 2 1 I I I I I I I
All other components of the contour must either only contribute to a separate part of the result (e.g. if I I I
There must be no singularities of our integrand along this contour.
The only singularities allowed within the region enclosed by the contour are poles.
One curve and integrand that would satisfy these criteria is the infinite semi-circle that covers the entire positive y y y y y y
with this contour integral:
∮ C z 2 e i m z ( z 4 + a 4 ) 2 d z . \oint_C \dfrac{z^2 e^{imz}}{(z^4+a^4)^2} dz. ∮ C ( z 4 + a 4 ) 2 z 2 e i m z d z . The residue theorem tells us that our contour integral will equal 2 π i ∑ r e s i d u e s 2\pi i \sum \mathrm{residues} 2 π i ∑ r e s i d u e s
z 4 + a 4 = 0 z 4 = − a 4 z 4 = a 4 ⋅ ( − 1 ) z n 4 = a 4 e π i + 2 π i n , n ∈ Z z n = a e π i ( 1 + 2 n ) 4 . \begin{aligned} z^4+a^4 &= 0 \\ z^4 &= -a^4 \\ z^4 &= a^4 \cdot (-1) \\ z^4_n &= a^4 e^{\pi i + 2\pi i n}, \hspace{0.1cm} n\in \mathbb{Z} \\ z_n &= a e^{\dfrac{\pi i(1+2n)}{4}}. \end{aligned} z 4 + a 4 z 4 z 4 z n 4 z n = 0 = − a 4 = a 4 ⋅ ( − 1 ) = a 4 e π i + 2 π i n , n ∈ Z = a e 4 π i ( 1 + 2 n ) . n = 0 , 1 , 2 , 3 n=0, 1, 2, 3 n = 0 , 1 , 2 , 3 z 4 + a 4 = ( z − z 0 ) ( z − z 1 ) ( z − z 2 ) ( z − z 3 ) z^4+a^4 = (z-z_0)(z-z_1)(z-z_2)(z-z_3) z 4 + a 4 = ( z − z 0 ) ( z − z 1 ) ( z − z 2 ) ( z − z 3 ) z 0 z_0 z 0 z n z_n z n n = 0 n=0 n = 0 z 1 z_1 z 1 ( z − z n ) 2 (z-z_n)^2 ( z − z n ) 2
The general formula for calculating residues is (where p p p k k k f ( z ) f(z) f ( z )
r e s i d u e = lim z → p 1 ( k − 1 ) ! d k − 1 d z k − 1 ( ( z − p ) k f ( z ) ) \begin{aligned} \mathrm{residue} = \lim_{z\rightarrow p} \dfrac{1}{(k-1)!} \dfrac{d^{k-1}}{dz^{k-1}} \left((z-p)^k f(z)\right) \end{aligned} r e s i d u e = z → p lim ( k − 1 ) ! 1 d z k − 1 d k − 1 ( ( z − p ) k f ( z ) ) For z 0 z_0 z 0 r 0 r_0 r 0
r 0 = lim z → z 0 1 ( 2 − 1 ) ! d 2 − 1 d z 2 − 1 ( ( z − z 0 ) 2 z 2 e i m z ( z 4 + a 4 ) 2 ) = lim z → z 0 d d z ( z 2 e i m z ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 ) = lim z → z 0 2 z e i m z ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 + i m z 2 e i m z ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 − 2 z 2 e i m z ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 ( 1 z − z 1 + 1 z − z 2 + 1 z − z 3 ) = lim z → z 0 z e i m z ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 ( 2 + i m z − 2 z ( 1 z − z 1 + 1 z − z 2 + 1 z − z 3 ) ) = z 0 e i m z 0 ( z 0 − z 1 ) 2 ( z 0 − z 2 ) 2 ( z 0 − z 3 ) 2 ( 2 + i m z 0 − 2 z 0 ( 1 z 0 − z 1 + 1 z 0 − z 2 + 1 z 0 − z 3 ) ) \begin{aligned} r_0 &= \lim_{z\rightarrow z_0} \dfrac{1}{(2-1)!} \dfrac{d^{2-1}}{dz^{2-1}} \left((z-z_0)^2 \dfrac{z^2e^{imz}}{(z^4+a^4)^2}\right) \\ &= \lim_{z\rightarrow z_0} \dfrac{d}{dz} \left(\dfrac{z^2 e^{imz}}{(z-z_1)^2(z-z_2)^2(z-z_3)^2}\right) \\ &= \lim_{z\rightarrow z_0} \dfrac{2ze^{imz}}{(z-z_1)^2(z-z_2)^2(z-z_3)^2} + \dfrac{imz^2 e^{imz}}{(z-z_1)^2(z-z_2)^2(z-z_3)^2} - \dfrac{2z^2e^{imz}}{(z-z_1)^2(z-z_2)^2(z-z_3)^2} \\\left(\dfrac{1}{z-z_1}+\dfrac{1}{z-z_2}+\dfrac{1}{z-z_3}\right) \\ &= \lim_{z\rightarrow z_0} \dfrac{ze^{imz}}{(z-z_1)^2(z-z_2)^2(z-z_3)^2} \left(2+imz-2z\left(\dfrac{1}{z-z_1}+\dfrac{1}{z-z_2}+\dfrac{1}{z-z_3}\right)\right) \\ &= \dfrac{z_0 e^{imz_0}}{(z_0-z_1)^2(z_0-z_2)^2(z_0-z_3)^2} \left(2 + imz_0 - 2z_0 \left(\dfrac{1}{z_0-z_1}+\dfrac{1}{z_0-z_2}+\dfrac{1}{z_0-z_3}\right)\right) \end{aligned} r 0 ( z − z 1 1 + z − z 2 1 + z − z 3 1 ) = z → z 0 lim ( 2 − 1 ) ! 1 d z 2 − 1 d 2 − 1 ( ( z − z 0 ) 2 ( z 4 + a 4 ) 2 z 2 e i m z ) = z → z 0 lim d z d ( ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 z 2 e i m z ) = z → z 0 lim ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 2 z e i m z + ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 i m z 2 e i m z − ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 2 z 2 e i m z = z → z 0 lim ( z − z 1 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 z e i m z ( 2 + i m z − 2 z ( z − z 1 1 + z − z 2 1 + z − z 3 1 ) ) = ( z 0 − z 1 ) 2 ( z 0 − z 2 ) 2 ( z 0 − z 3 ) 2 z 0 e i m z 0 ( 2 + i m z 0 − 2 z 0 ( z 0 − z 1 1 + z 0 − z 2 1 + z 0 − z 3 1 ) ) To simplify our above equation, we need to simplify our denominators:
z 0 − z 1 = a ( e π i 4 − e 3 π i 4 ) = a e π i 4 ( 1 − e π i 2 ) = a e π i 4 ( 1 − i ) = a e π i 4 2 e − π i 4 = a 2 . z 0 − z 2 = a ( e π i 4 − e 5 π i 4 ) = a e π i 4 ( 1 − e π i ) = a e π i 4 ( 1 − ( − 1 ) ) = 2 a e π i 4 . z 0 − z 3 = a ( e π i 4 − e 7 π i 4 ) = a e π i 4 ( 1 − e 3 π i 2 ) = a e π i 4 ( 1 − ( − i ) ) = a e π i 4 ( 1 + i ) = a e π i 4 2 e π i 4 = a 2 e π i 2 = a 2 i . \begin{aligned} z_0-z_1 &= a \left(e^{\dfrac{\pi i}{4}} - e^{\dfrac{3\pi i}{4}}\right) \\ &= ae^{\dfrac{\pi i}{4}} \left(1-e^{\dfrac{\pi i}{2}}\right) \\ &= ae^{\dfrac{\pi i}{4}} (1-i) \\ &= ae^{\dfrac{\pi i}{4}} \sqrt{2} e^{-\dfrac{\pi i}{4}} \\ &= a\sqrt{2}.\\ z_0 - z_2 &= a \left(e^{\dfrac{\pi i}{4}} - e^{\dfrac{5\pi i}{4}}\right) \\ &= ae^{\dfrac{\pi i}{4}} \left(1-e^{\pi i}\right) \\ &= ae^{\dfrac{\pi i}{4}} (1 - (-1)) \\ &= 2ae^{\dfrac{\pi i}{4}}.\\ z_0 - z_3 &= a \left(e^{\dfrac{\pi i}{4}} - e^{\dfrac{7\pi i}{4}}\right) \\ &= ae^{\dfrac{\pi i}{4}} \left(1-e^{\dfrac{3\pi i}{2}}\right) \\ &= ae^{\dfrac{\pi i}{4}} (1 - (-i))\\ &= ae^{\dfrac{\pi i}{4}} (1+i) \\ &= ae^{\dfrac{\pi i}{4}} \sqrt{2} e^{\dfrac{\pi i}{4}} \\ &= a\sqrt{2} e^{\dfrac{\pi i}{2}} \\ &= a\sqrt{2} i. \end{aligned} z 0 − z 1 z 0 − z 2 z 0 − z 3 = a ⎝ ⎜ ⎜ ⎛ e 4 π i − e 4 3 π i ⎠ ⎟ ⎟ ⎞ = a e 4 π i ⎝ ⎜ ⎜ ⎛ 1 − e 2 π i ⎠ ⎟ ⎟ ⎞ = a e 4 π i ( 1 − i ) = a e 4 π i 2 e − 4 π i = a 2 . = a ⎝ ⎜ ⎜ ⎛ e 4 π i − e 4 5 π i ⎠ ⎟ ⎟ ⎞ = a e 4 π i ( 1 − e π i ) = a e 4 π i ( 1 − ( − 1 ) ) = 2 a e 4 π i . = a ⎝ ⎜ ⎜ ⎛ e 4 π i − e 4 7 π i ⎠ ⎟ ⎟ ⎞ = a e 4 π i ⎝ ⎜ ⎜ ⎛ 1 − e 2 3 π i ⎠ ⎟ ⎟ ⎞ = a e 4 π i ( 1 − ( − i ) ) = a e 4 π i ( 1 + i ) = a e 4 π i 2 e 4 π i = a 2 e 2 π i = a 2 i . Likewise, i z 0 = e π i 2 ⋅ a e π i 4 = a e 3 π i 4 = a 2 ( − 1 + i ) iz_0 = e^{\dfrac{\pi i}{2}} \cdot ae^{\dfrac{\pi i}{4}} = ae^{\dfrac{3\pi i}{4}} = \dfrac{a}{\sqrt{2}} (-1+i) i z 0 = e 2 π i ⋅ a e 4 π i = a e 4 3 π i = 2 a ( − 1 + i )
r 0 = a e π i 4 e a m 2 ( − 1 + i ) 2 a 2 ⋅ 4 a 2 e π i 2 ⋅ 2 a 2 e π i ( 2 + a m 2 ( − 1 + i ) − 2 a e π i 4 ( 1 a 2 + 1 2 a e π i 4 + 1 a 2 e π i 2 ) ) = e − 5 π i 4 e a m 2 ( − 1 + i ) 16 a 5 ( 2 + a m 2 ( − 1 + i ) − 2 e π i 4 − 1 − 2 e − π i 4 ) \begin{aligned} r_0 &= \dfrac{ae^{\dfrac{\pi i}{4}} e^{\dfrac{am}{\sqrt{2}}(-1+i)}}{2a^2 \cdot 4a^2e^{\dfrac{\pi i}{2}} \cdot 2a^2 e^{\pi i}} \left(2+\dfrac{am}{\sqrt{2}}(-1+i) - 2ae^{\dfrac{\pi i}{4}}\left(\dfrac{1}{a\sqrt{2}} + \dfrac{1}{2ae^{\dfrac{\pi i}{4}}} + \dfrac{1}{a\sqrt{2}e^{\dfrac{\pi i}{2}}}\right)\right) \\ &= \dfrac{e^{-\dfrac{5\pi i}{4}} e^{\dfrac{am}{\sqrt{2}}(-1+i)}}{16a^5} \left(2+\dfrac{am}{\sqrt{2}}(-1+i) - \sqrt{2}e^{\dfrac{\pi i}{4}} - 1 - \sqrt{2}e^{-\dfrac{\pi i}{4}}\right) \\ \end{aligned} r 0 = 2 a 2 ⋅ 4 a 2 e 2 π i ⋅ 2 a 2 e π i a e 4 π i e 2 a m ( − 1 + i ) ⎝ ⎜ ⎜ ⎜ ⎛ 2 + 2 a m ( − 1 + i ) − 2 a e 4 π i ⎝ ⎜ ⎜ ⎜ ⎛ a 2 1 + 2 a e 4 π i 1 + a 2 e 2 π i 1 ⎠ ⎟ ⎟ ⎟ ⎞ ⎠ ⎟ ⎟ ⎟ ⎞ = 1 6 a 5 e − 4 5 π i e 2 a m ( − 1 + i ) ⎝ ⎜ ⎜ ⎛ 2 + 2 a m ( − 1 + i ) − 2 e 4 π i − 1 − 2 e − 4 π i ⎠ ⎟ ⎟ ⎞ Using: e π i 4 + e − π i 4 = 2 e^{\dfrac{\pi i}{4}} + e^{-\dfrac{\pi i}{4}} = \sqrt{2} e 4 π i + e − 4 π i = 2 e − 5 π i 4 = e 3 π i 4 e^{-\dfrac{5\pi i}{4}} = e^{\dfrac{3\pi i}{4}} e − 4 5 π i = e 4 3 π i
r 0 = e 3 π i 4 e a m 2 ( − 1 + i ) 16 a 5 ( a m 2 ( − 1 + i ) − 1 ) = e a m 2 ( − 1 + i ) 16 a 5 2 ( a m 2 ( − 1 + i ) 2 − ( − 1 + i ) ) = e a m 2 ( − 1 + i ) 16 a 5 2 ( a m 2 ⋅ − 2 i + 1 − i ) = e a m 2 ( − 1 + i ) 16 a 5 2 ( − i ( a m 2 + 1 ) + 1 ) \begin{aligned} r_0 &= \dfrac{e^{\dfrac{3\pi i}{4}} e^{\dfrac{am}{\sqrt{2}}(-1+i)}}{16a^5} \left(\dfrac{am}{\sqrt{2}}(-1+i) -1\right) \\ &= \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1+i)}}{16a^5\sqrt{2}}\left(\dfrac{am}{\sqrt{2}}(-1+i)^2 - (-1+i)\right) \\ &= \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1+i)}}{16a^5\sqrt{2}}\left(\dfrac{am}{\sqrt{2}}\cdot -2i+1-i\right) \\ &= \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1+i)}}{16a^5\sqrt{2}}\left(-i\left(am\sqrt{2}+1\right)+1\right) \end{aligned} r 0 = 1 6 a 5 e 4 3 π i e 2 a m ( − 1 + i ) ( 2 a m ( − 1 + i ) − 1 ) = 1 6 a 5 2 e 2 a m ( − 1 + i ) ( 2 a m ( − 1 + i ) 2 − ( − 1 + i ) ) = 1 6 a 5 2 e 2 a m ( − 1 + i ) ( 2 a m ⋅ − 2 i + 1 − i ) = 1 6 a 5 2 e 2 a m ( − 1 + i ) ( − i ( a m 2 + 1 ) + 1 ) Next, we must find r 1 r_1 r 1
r 1 = lim z → z 1 d d z ( ( z − z 1 ) 2 z 2 e i m z ( z 4 + a 4 ) 2 ) = lim z → z 1 d d z ( z 2 e i m z ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 ) = lim z → z 1 2 z e i m z ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 + i m z 2 e i m z ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 − 2 z 2 e i m z ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 ( 1 z − z 0 + 1 z − z 2 + 1 z − z 3 ) = lim z → z 1 z e i m z ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 ( 2 + i m z − 2 z ( 1 z − z 0 + 1 z − z 2 + 1 z − z 3 ) ) = z 1 e i m z 1 ( z 1 − z 0 ) 2 ( z 1 − z 2 ) 2 ( z 1 − z 3 ) 2 ( 2 + i m z 1 − 2 z 1 ( 1 z 1 − z 0 + 1 z 1 − z 2 + 1 z 1 − z 3 ) ) . \begin{aligned} r_1 &= \lim_{z\rightarrow z_1} \dfrac{d}{dz} \left((z-z_1)^2 \dfrac{z^2e^{imz}}{(z^4+a^4)^2}\right) \\ &= \lim_{z\rightarrow z_1} \dfrac{d}{dz} \left(\dfrac{z^2 e^{imz}}{(z-z_0)^2(z-z_2)^2(z-z_3)^2}\right) \\ &= \lim_{z\rightarrow z_1} \dfrac{2ze^{imz}}{(z-z_0)^2(z-z_2)^2(z-z_3)^2} + \dfrac{imz^2 e^{imz}}{(z-z_0)^2(z-z_2)^2(z-z_3)^2} - \dfrac{2z^2e^{imz}}{(z-z_0)^2(z-z_2)^2(z-z_3)^2} \\\left(\dfrac{1}{z-z_0}+\dfrac{1}{z-z_2}+\dfrac{1}{z-z_3}\right) \\ &= \lim_{z\rightarrow z_1} \dfrac{ze^{imz}}{(z-z_0)^2(z-z_2)^2(z-z_3)^2} \left(2+imz-2z\left(\dfrac{1}{z-z_0}+\dfrac{1}{z-z_2}+\dfrac{1}{z-z_3}\right)\right) \\ &= \dfrac{z_1 e^{imz_1}}{(z_1-z_0)^2(z_1-z_2)^2(z_1-z_3)^2} \left(2 + imz_1 - 2z_1 \left(\dfrac{1}{z_1-z_0}+\dfrac{1}{z_1-z_2}+\dfrac{1}{z_1-z_3}\right)\right). \end{aligned} r 1 ( z − z 0 1 + z − z 2 1 + z − z 3 1 ) = z → z 1 lim d z d ( ( z − z 1 ) 2 ( z 4 + a 4 ) 2 z 2 e i m z ) = z → z 1 lim d z d ( ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 z 2 e i m z ) = z → z 1 lim ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 2 z e i m z + ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 i m z 2 e i m z − ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 2 z 2 e i m z = z → z 1 lim ( z − z 0 ) 2 ( z − z 2 ) 2 ( z − z 3 ) 2 z e i m z ( 2 + i m z − 2 z ( z − z 0 1 + z − z 2 1 + z − z 3 1 ) ) = ( z 1 − z 0 ) 2 ( z 1 − z 2 ) 2 ( z 1 − z 3 ) 2 z 1 e i m z 1 ( 2 + i m z 1 − 2 z 1 ( z 1 − z 0 1 + z 1 − z 2 1 + z 1 − z 3 1 ) ) . First, we shall find z 1 − z 0 z_1-z_0 z 1 − z 0 z 1 − z 2 z_1-z_2 z 1 − z 2 z 1 − z 3 z_1-z_3 z 1 − z 3
z 1 − z 0 = − ( z 0 − z 1 ) = − a 2 . z 1 − z 2 = a ( e 3 π i 4 − e 5 π i 4 ) = a e 3 π i 4 ( 1 − e π i 2 ) = a e 3 π i 4 ( 1 − i ) = a e 3 π i 4 ⋅ 2 e − π i 4 = a 2 e π i 2 . z 1 − z 3 = a ( e 3 π i 4 − e 7 π i 4 ) = a e 3 π i 4 ( 1 − e π i ) = a e 3 π i 4 ( 1 − ( − 1 ) ) = 2 a e 3 π i 4 . i z 1 = e π i 2 a e 3 π i 4 = a e 5 π i 4 = a 2 ( − 1 − i ) . \begin{aligned} z_1 - z_0 &= -(z_0-z_1) \\ &= -a\sqrt{2}.\\ z_1 - z_2 &= a \left(e^{\dfrac{3\pi i}{4}}-e^{\dfrac{5\pi i}{4}}\right) \\ &= ae^{\dfrac{3\pi i}{4}}(1-e^{\dfrac{\pi i}{2}}) \\ &= ae^{\dfrac{3\pi i}{4}}(1-i) \\ &= ae^{\dfrac{3\pi i}{4}} \cdot \sqrt{2} e^{-\dfrac{\pi i}{4}} \\ &= a\sqrt{2} e^{\dfrac{\pi i}{2}}.\\ z_1 - z_3 &= a \left(e^{\dfrac{3\pi i}{4}}-e^{\dfrac{7\pi i}{4}}\right) \\ &= ae^{\dfrac{3\pi i}{4}} (1-e^{\pi i}) \\ &= ae^{\dfrac{3\pi i}{4}} (1 - (-1)) \\ &= 2a e^{\dfrac{3\pi i}{4}}. \\ iz_1 &= e^{\dfrac{\pi i}{2}} ae^{\dfrac{3\pi i}{4}} \\ &= ae^{\dfrac{5\pi i}{4}} \\ &= \dfrac{a}{\sqrt{2}} (-1-i). \end{aligned} z 1 − z 0 z 1 − z 2 z 1 − z 3 i z 1 = − ( z 0 − z 1 ) = − a 2 . = a ⎝ ⎜ ⎜ ⎛ e 4 3 π i − e 4 5 π i ⎠ ⎟ ⎟ ⎞ = a e 4 3 π i ( 1 − e 2 π i ) = a e 4 3 π i ( 1 − i ) = a e 4 3 π i ⋅ 2 e − 4 π i = a 2 e 2 π i . = a ⎝ ⎜ ⎜ ⎛ e 4 3 π i − e 4 7 π i ⎠ ⎟ ⎟ ⎞ = a e 4 3 π i ( 1 − e π i ) = a e 4 3 π i ( 1 − ( − 1 ) ) = 2 a e 4 3 π i . = e 2 π i a e 4 3 π i = a e 4 5 π i = 2 a ( − 1 − i ) . r 1 = a e 3 π i 4 e a m 2 ( − 1 − i ) 2 a 2 ⋅ 2 a 2 e π i ⋅ 4 a 2 e 3 π i 2 ( 2 + a m 2 ( − 1 − i ) − 2 a e 3 π i 4 ( − 1 a 2 + 1 a 2 e π i 2 + 1 2 a e 3 π i 4 ) ) = e − 7 π i 4 e a m 2 ( − 1 − i ) 16 a 5 ( 2 − a m 2 ( 1 + i ) + 2 e 3 π i 4 − 2 e π i 4 − 1 ) \begin{aligned} r_1 &= \dfrac{ae^{\dfrac{3\pi i}{4}}e^{\dfrac{am}{\sqrt{2}}(-1-i)}}{2a^2 \cdot 2a^2 e^{\pi i} \cdot 4a^2 e^{\dfrac{3\pi i}{2}}} \left(2 + \dfrac{am}{\sqrt{2}}(-1-i)-2ae^{\dfrac{3\pi i}{4}}\left(-\dfrac{1}{a\sqrt{2}}+\dfrac{1}{a\sqrt{2}e^{\dfrac{\pi i}{2}}}+\dfrac{1}{2ae^{\dfrac{3\pi i}{4}}}\right)\right) \\ &= \dfrac{e^{-\dfrac{7\pi i}{4}e^{\frac{am}{\sqrt{2}}(-1-i)}}}{16a^5} (2-\dfrac{am}{\sqrt{2}}(1+i)+\sqrt{2}e^{\dfrac{3\pi i}{4}} - \sqrt{2} e^{\dfrac{\pi i}{4}}-1) \end{aligned} r 1 = 2 a 2 ⋅ 2 a 2 e π i ⋅ 4 a 2 e 2 3 π i a e 4 3 π i e 2 a m ( − 1 − i ) ⎝ ⎜ ⎜ ⎜ ⎛ 2 + 2 a m ( − 1 − i ) − 2 a e 4 3 π i ⎝ ⎜ ⎜ ⎜ ⎛ − a 2 1 + a 2 e 2 π i 1 + 2 a e 4 3 π i 1 ⎠ ⎟ ⎟ ⎟ ⎞ ⎠ ⎟ ⎟ ⎟ ⎞ = 1 6 a 5 e − 4 7 π i e 2 a m ( − 1 − i ) ( 2 − 2 a m ( 1 + i ) + 2 e 4 3 π i − 2 e 4 π i − 1 ) To simplify the above expression, let us do some complex arithmetic:
2 e 3 π i 4 − 2 e π i 4 = 2 ⋅ 1 2 ( − 1 + i − 1 − i ) = − 2. e − 7 π i 4 = e π i 4 = 1 2 ( 1 + i ) . \begin{aligned} \sqrt{2} e^{\dfrac{3\pi i}{4}} - \sqrt{2}e^{\dfrac{\pi i}{4}} &= \sqrt{2} \cdot \dfrac{1}{\sqrt{2}} (-1+i - 1 - i) \\ &= -2.\\ e^{-\dfrac{7\pi i}{4}} &= e^{\dfrac{\pi i}{4}} \\ &= \dfrac{1}{\sqrt{2}}(1+i). \end{aligned} 2 e 4 3 π i − 2 e 4 π i e − 4 7 π i = 2 ⋅ 2 1 ( − 1 + i − 1 − i ) = − 2 . = e 4 π i = 2 1 ( 1 + i ) . r 1 = e a m 2 ( − 1 − i ) ( 1 + i ) 16 a 5 2 ( − a m 2 ( 1 + i ) − 1 ) = e a m 2 ( − 1 − i ) 16 a 5 2 ( − a m 2 i − 1 − i ) = e a m 2 ( − 1 − i ) 16 a 5 2 ( − i ( 1 + a m 2 ) − 1 ) \begin{aligned} r_1 &= \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1-i)}(1+i)}{16a^5 \sqrt{2}}\left(-\dfrac{am}{\sqrt{2}}(1+i)-1\right) \\ &= \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1-i)}}{16a^5 \sqrt{2}}\left(-am\sqrt{2}i - 1 - i\right) \\ &= \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1-i)}}{16a^5 \sqrt{2}}\left(-i(1+am\sqrt{2})-1\right) \end{aligned} r 1 = 1 6 a 5 2 e 2 a m ( − 1 − i ) ( 1 + i ) ( − 2 a m ( 1 + i ) − 1 ) = 1 6 a 5 2 e 2 a m ( − 1 − i ) ( − a m 2 i − 1 − i ) = 1 6 a 5 2 e 2 a m ( − 1 − i ) ( − i ( 1 + a m 2 ) − 1 ) The sum of the residues is thus:
∑ r e s i d u e s = e a m 2 ( − 1 + i ) 16 a 5 2 ( − i ( a m 2 + 1 ) + 1 ) + e a m 2 ( − 1 − i ) 16 a 5 2 ( − i ( 1 + a m 2 ) − 1 ) = e − a m 2 16 a 5 2 ( e a m i 2 − e − a m i 2 − i ( a m 2 + 1 ) ( e a m i 2 + e − a m i 2 ) ) = e − a m 2 16 a 5 2 ( 2 i sin a m 2 − 2 i ( a m 2 + 1 ) cos a m 2 ) = e − a m 2 i 8 a 5 2 ( sin a m 2 − ( a m 2 + 1 ) cos a m 2 ) \begin{aligned} \sum \mathrm{residues} &= \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1+i)}}{16a^5\sqrt{2}}\left(-i\left(am\sqrt{2}+1\right)+1\right) + \dfrac{e^{\dfrac{am}{\sqrt{2}}(-1-i)}}{16a^5 \sqrt{2}}\left(-i(1+am\sqrt{2})-1\right)\\ &= \dfrac{e^{-\dfrac{am}{\sqrt{2}}}}{16a^5\sqrt{2}} \left(e^{\dfrac{ami}{\sqrt{2}}}-e^{-\dfrac{ami}{\sqrt{2}}} -i (am\sqrt{2}+1)\left(e^{\dfrac{ami}{\sqrt{2}}}+e^{-\dfrac{ami}{\sqrt{2}}}\right)\right) \\ &= \dfrac{e^{-\dfrac{am}{\sqrt{2}}}}{16a^5\sqrt{2}} \left(2i\sin{\dfrac{am}{\sqrt{2}}} -2i (am\sqrt{2}+1)\cos{\dfrac{am}{\sqrt{2}}}\right) \\ &= \dfrac{e^{-\dfrac{am}{\sqrt{2}}}i}{8a^5\sqrt{2}}\left(\sin{\dfrac{am}{\sqrt{2}}} -(am\sqrt{2}+1)\cos{\dfrac{am}{\sqrt{2}}}\right) \\ \end{aligned} ∑ r e s i d u e s = 1 6 a 5 2 e 2 a m ( − 1 + i ) ( − i ( a m 2 + 1 ) + 1 ) + 1 6 a 5 2 e 2 a m ( − 1 − i ) ( − i ( 1 + a m 2 ) − 1 ) = 1 6 a 5 2 e − 2 a m ⎝ ⎜ ⎜ ⎛ e 2 a m i − e − 2 a m i − i ( a m 2 + 1 ) ⎝ ⎜ ⎜ ⎛ e 2 a m i + e − 2 a m i ⎠ ⎟ ⎟ ⎞ ⎠ ⎟ ⎟ ⎞ = 1 6 a 5 2 e − 2 a m ( 2 i sin 2 a m − 2 i ( a m 2 + 1 ) cos 2 a m ) = 8 a 5 2 e − 2 a m i ( sin 2 a m − ( a m 2 + 1 ) cos 2 a m ) Therefore our contour integral equals:
∮ C z 2 e i m z ( z 4 + a 4 ) 2 d z = 2 π i ∑ r e s i d u e s = e − a m 2 π 4 a 5 2 ( ( a m 2 + 1 ) cos a m 2 − sin a m 2 ) . (1) \begin{aligned} \oint_C \dfrac{z^2 e^{imz}}{(z^4+a^4)^2} dz &= 2 \pi i \sum \mathrm{residues} \\ &= \dfrac{e^{-\dfrac{am}{\sqrt{2}}}\pi}{4a^5\sqrt{2}}\left((am\sqrt{2}+1)\cos{\dfrac{am}{\sqrt{2}}}-\sin{\dfrac{am}{\sqrt{2}}}\right).\hspace{5cm}\tag{1} \end{aligned} ∮ C ( z 4 + a 4 ) 2 z 2 e i m z d z = 2 π i ∑ r e s i d u e s = 4 a 5 2 e − 2 a m π ( ( a m 2 + 1 ) cos 2 a m − sin 2 a m ) . ( 1 ) Breaking our contour integral into its components:
∮ C z 2 e i m z ( z 4 + a 4 ) 2 d z = ∫ A B z 2 e i m z ( z 4 + a 4 ) 2 d z + ∫ B A z 2 e i m z ( z 4 + a 4 ) 2 d z \begin{aligned} \oint_C \dfrac{z^2 e^{imz}}{(z^4+a^4)^2} dz &= \int_{AB} \dfrac{z^2 e^{imz}}{(z^4+a^4)^2} dz + \int_{BA} \dfrac{z^2 e^{imz}}{(z^4+a^4)^2} dz \end{aligned} ∮ C ( z 4 + a 4 ) 2 z 2 e i m z d z = ∫ A B ( z 4 + a 4 ) 2 z 2 e i m z d z + ∫ B A ( z 4 + a 4 ) 2 z 2 e i m z d z Along AB: z = x z=x z = x d z = d x dz=dx d z = d x x x x − R -R − R R R R z = R e i θ z=Re^{i\theta} z = R e i θ d z = i R e i θ d θ dz=iRe^{i\theta}d\theta d z = i R e i θ d θ θ \theta θ 0 0 0 π \pi π
∫ A B z 2 e i m z ( z 4 + a 4 ) 2 d z + ∫ B A z 2 e i m z ( z 4 + a 4 ) 2 d z = ∫ − R R x 2 e i m x ( x 4 + a 4 ) 2 d x + ∫ 0 π R 2 e 2 i θ e i m R e i θ ( R 4 e 4 i θ + a 4 ) 2 i R e i θ d θ = ∫ − R R x 2 e i m x ( x 4 + a 4 ) 2 d x + i ∫ 0 π R 3 e 3 i θ e i m R e i θ ( R 4 e 4 i θ + a 4 ) 2 d θ . \begin{aligned} \int_{AB} \dfrac{z^2 e^{imz}}{(z^4+a^4)^2} dz + \int_{BA} \dfrac{z^2 e^{imz}}{(z^4+a^4)^2} dz &= \int_{-R}^R \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + \int_{0}^{\pi} \dfrac{R^2 e^{2i\theta} e^{imRe^{i\theta}}}{(R^4e^{4i\theta}+a^4)^2} iRe^{i\theta} d\theta \\ &= \int_{-R}^R \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + i\int_{0}^{\pi} \dfrac{R^3 e^{3i\theta} e^{imRe^{i\theta}}}{(R^4e^{4i\theta}+a^4)^2} d\theta. \\ \end{aligned} ∫ A B ( z 4 + a 4 ) 2 z 2 e i m z d z + ∫ B A ( z 4 + a 4 ) 2 z 2 e i m z d z = ∫ − R R ( x 4 + a 4 ) 2 x 2 e i m x d x + ∫ 0 π ( R 4 e 4 i θ + a 4 ) 2 R 2 e 2 i θ e i m R e i θ i R e i θ d θ = ∫ − R R ( x 4 + a 4 ) 2 x 2 e i m x d x + i ∫ 0 π ( R 4 e 4 i θ + a 4 ) 2 R 3 e 3 i θ e i m R e i θ d θ . Taking the limit as R → ∞ R\rightarrow\infty R → ∞
lim R → ∞ ∫ − R R x 2 e i m x ( x 4 + a 4 ) 2 d x + i ∫ 0 π R 3 e 3 i θ e i m R e i θ ( R 4 e 4 i θ + a 4 ) 2 d θ = ∫ − ∞ ∞ x 2 e i m x ( x 4 + a 4 ) 2 d x + lim R → ∞ i ∫ 0 π R 3 e 3 i θ e i m R e i θ ( R 4 e 4 i θ + a 4 ) 2 d θ = ∫ − ∞ ∞ x 2 e i m x ( x 4 + a 4 ) 2 d x + lim R → ∞ i ∫ 0 π R 3 e 3 i θ e i m R e i θ R 8 e 8 i θ + a 8 + 2 a 4 R 4 e 4 i θ d θ = ∫ − ∞ ∞ x 2 e i m x ( x 4 + a 4 ) 2 d x + lim R → ∞ i ∫ 0 π e i m R e i θ R 5 e 5 i θ + a 8 e − 3 i θ R − 3 + 2 a 4 R e i θ d θ = ∫ − ∞ ∞ x 2 e i m x ( x 4 + a 4 ) 2 d x + lim R → ∞ i ∫ 0 π e i m R cos θ − m R sin θ R 5 e 5 i θ + a 8 e − 3 i θ R − 3 + 2 a 4 R e i θ d θ = ∫ − ∞ ∞ x 2 e i m x ( x 4 + a 4 ) 2 d x + lim R → ∞ i ∫ 0 π e i m R cos θ e − m R sin θ R 5 e 5 i θ + a 8 e − 3 i θ R − 3 + 2 a 4 R e i θ d θ \begin{aligned} \lim_{R\rightarrow \infty} \int_{-R}^R \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + i\int_{0}^{\pi} \dfrac{R^3 e^{3i\theta} e^{imRe^{i\theta}}}{(R^4e^{4i\theta}+a^4)^2} d\theta &= \int_{-\infty}^{\infty} \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + \lim_{R\rightarrow \infty} i\int_{0}^{\pi} \dfrac{R^3 e^{3i\theta} e^{imRe^{i\theta}}}{(R^4e^{4i\theta}+a^4)^2} d\theta \\ &= \int_{-\infty}^{\infty} \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + \lim_{R\rightarrow \infty} i\int_{0}^{\pi} \dfrac{R^3 e^{3i\theta} e^{imRe^{i\theta}}}{R^8e^{8i\theta}+a^8+2a^4R^4e^{4i\theta}} d\theta \\ &= \int_{-\infty}^{\infty} \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + \lim_{R\rightarrow \infty} i\int_{0}^{\pi} \dfrac{e^{imRe^{i\theta}}}{R^5e^{5i\theta}+a^8e^{-3i\theta}R^{-3}+2a^4Re^{i\theta}} d\theta \\ &= \int_{-\infty}^{\infty} \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + \lim_{R\rightarrow \infty} i\int_{0}^{\pi} \dfrac{e^{imR\cos{\theta}-mR\sin{\theta}}}{R^5e^{5i\theta}+a^8e^{-3i\theta}R^{-3}+2a^4Re^{i\theta}} d\theta \\ &= \int_{-\infty}^{\infty} \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx + \lim_{R\rightarrow \infty} i\int_{0}^{\pi} \dfrac{e^{imR\cos{\theta}}e^{-mR\sin{\theta}}}{R^5e^{5i\theta}+a^8e^{-3i\theta}R^{-3}+2a^4Re^{i\theta}} d\theta \end{aligned} R → ∞ lim ∫ − R R ( x 4 + a 4 ) 2 x 2 e i m x d x + i ∫ 0 π ( R 4 e 4 i θ + a 4 ) 2 R 3 e 3 i θ e i m R e i θ d θ = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 e i m x d x + R → ∞ lim i ∫ 0 π ( R 4 e 4 i θ + a 4 ) 2 R 3 e 3 i θ e i m R e i θ d θ = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 e i m x d x + R → ∞ lim i ∫ 0 π R 8 e 8 i θ + a 8 + 2 a 4 R 4 e 4 i θ R 3 e 3 i θ e i m R e i θ d θ = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 e i m x d x + R → ∞ lim i ∫ 0 π R 5 e 5 i θ + a 8 e − 3 i θ R − 3 + 2 a 4 R e i θ e i m R e i θ d θ = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 e i m x d x + R → ∞ lim i ∫ 0 π R 5 e 5 i θ + a 8 e − 3 i θ R − 3 + 2 a 4 R e i θ e i m R c o s θ − m R s i n θ d θ = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 e i m x d x + R → ∞ lim i ∫ 0 π R 5 e 5 i θ + a 8 e − 3 i θ R − 3 + 2 a 4 R e i θ e i m R c o s θ e − m R s i n θ d θ Within the domain of integration for the second integral, sin θ ≥ 0 \sin{\theta} \geq 0 sin θ ≥ 0 e − m R sin θ ≤ 1 e^{-mR\sin{\theta}} \leq 1 e − m R s i n θ ≤ 1 m ≥ 0 m\geq0 m ≥ 0 R R R R → ∞ R\rightarrow\infty R → ∞
Therefore our contour integral becomes:
∫ − ∞ ∞ x 2 e i m x ( x 4 + a 4 ) 2 d x = ∫ − ∞ ∞ x 2 ( cos m x + i sin m x ) ( x 4 + a 4 ) 2 d x = ∫ − ∞ ∞ x 2 cos m x ( x 4 + a 4 ) 2 d x + i ∫ − ∞ ∞ x 2 sin m x ( x 4 + a 4 ) 2 d x \begin{aligned} \int_{-\infty}^{\infty} \dfrac{x^2 e^{imx}}{(x^4+a^4)^2} dx &= \int_{-\infty}^{\infty} \dfrac{x^2 (\cos{mx}+i\sin{mx})}{(x^4+a^4)^2} dx \\ &= \int_{-\infty}^{\infty} \dfrac{x^2 \cos{mx}}{(x^4+a^4)^2} dx + i \int_{-\infty}^{\infty} \dfrac{x^2 \sin{mx}}{(x^4+a^4)^2} dx \end{aligned} ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 e i m x d x = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 ( cos m x + i sin m x ) d x = ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 cos m x d x + i ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 sin m x d x Equating with the right-hand side of (1):
∫ − ∞ ∞ x 2 cos m x ( x 4 + a 4 ) 2 d x + i ∫ − ∞ ∞ x 2 sin m x ( x 4 + a 4 ) 2 d x = e − a m 2 π 4 a 5 2 ( ( a m 2 + 1 ) cos a m 2 − sin a m 2 ) \begin{aligned} \int_{-\infty}^{\infty} \dfrac{x^2 \cos{mx}}{(x^4+a^4)^2} dx + i \int_{-\infty}^{\infty} \dfrac{x^2 \sin{mx}}{(x^4+a^4)^2} dx &= \dfrac{e^{-\dfrac{am}{\sqrt{2}}}\pi}{4a^5\sqrt{2}}\left((am\sqrt{2}+1)\cos{\dfrac{am}{\sqrt{2}}}-\sin{\dfrac{am}{\sqrt{2}}}\right) \end{aligned} ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 cos m x d x + i ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 sin m x d x = 4 a 5 2 e − 2 a m π ( ( a m 2 + 1 ) cos 2 a m − sin 2 a m ) Equating real and imaginary parts:
∫ − ∞ ∞ x 2 cos m x ( x 4 + a 4 ) 2 d x = e − a m 2 π 4 a 5 2 ( ( a m 2 + 1 ) cos a m 2 − sin a m 2 ) ∫ − ∞ ∞ x 2 sin m x ( x 4 + a 4 ) 2 d x = 0. \begin{aligned} \int_{-\infty}^{\infty} \dfrac{x^2 \cos{mx}}{(x^4+a^4)^2} dx &= \dfrac{e^{-\dfrac{am}{\sqrt{2}}}\pi}{4a^5\sqrt{2}}\left((am\sqrt{2}+1)\cos{\dfrac{am}{\sqrt{2}}}-\sin{\dfrac{am}{\sqrt{2}}}\right) \\ \int_{-\infty}^{\infty} \dfrac{x^2 \sin{mx}}{(x^4+a^4)^2} dx &= 0. \end{aligned} ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 cos m x d x ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 sin m x d x = 4 a 5 2 e − 2 a m π ( ( a m 2 + 1 ) cos 2 a m − sin 2 a m ) = 0 . It is important to note, however, that because cos ( − m x ) = cos m x \cos{(-mx)}=\cos{mx} cos ( − m x ) = cos m x ( − a ) 4 = a 4 (-a)^4=a^4 ( − a ) 4 = a 4 a a a m m m a a a m m m
∫ − ∞ ∞ x 2 cos m x ( x 4 + a 4 ) 2 d x = e − ∣ a m ∣ 2 π 4 ∣ a ∣ 5 2 ( ( ∣ a m ∣ 2 + 1 ) cos a m 2 − sin ∣ a m ∣ 2 ) \int_{-\infty}^{\infty} \dfrac{x^2 \cos{mx}}{(x^4+a^4)^2} dx = \dfrac{e^{-\dfrac{|am|}{\sqrt{2}}}\pi}{4|a|^5\sqrt{2}}\left((|am|\sqrt{2}+1)\cos{\dfrac{am}{\sqrt{2}}}-\sin{\dfrac{|am|}{\sqrt{2}}}\right) ∫ − ∞ ∞ ( x 4 + a 4 ) 2 x 2 cos m x d x = 4 ∣ a ∣ 5 2 e − 2 ∣ a m ∣ π ( ( ∣ a m ∣ 2 + 1 ) cos 2 a m − sin 2 ∣ a m ∣ ) 