Evaluating the integral of sine mx over x times x to the eighth plus a to the eighth

The integral we wish to evaluate is I=sinmxx(x8+a8)dxI = \displaystyle \int_{-\infty}^{\infty} \dfrac{\sin{mx}}{x(x^8+a^8)} dx, where a,m>0a, m \gt 0 and are real.

Curve and integrand hunting

The best curve to use to evaluate this integral is likely this one:

We will use this contour integral: Ceimzz(z8+a8)dz\displaystyle \oint_C \dfrac{e^{imz}}{z(z^8+a^8)} dz.

Pole hunting and residue evaluation

The integrand of which has poles where:

z8+a8=0z8=a8z8=a8(1)zn8=a8eπi(1+2n),nZzn=aeπi8(1+2n). \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}

n=0,1,2,3,4,5,6,7n=0,1,2,3,4,5,6,7 gives us our only unique poles (which are all first order). Of these, only n=0,1,2,3n=0,1,2,3 is within the region enclosed by C. Let rnr_n be the residue of our integrand at znz_n.

rn=limzzneimz(zz0)z(z8+a8)=limzzneimzz8z7=eimzn8zn8=eimzn8a8. \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}

Let us do some working to simplify r0=eimz08a8.r_0 = -\dfrac{e^{imz_0}}{8a^8}. imz0=ame5πi8=amsinπ8+amicosπ8imz_0 = ame^{\dfrac{5\pi i}{8}} = -am\sin{\dfrac{\pi}{8}}+ami\cos{\dfrac{\pi}{8}}. Let p=sinπ8p=\sin{\dfrac{\pi}{8}} and q=cosπ8q=\cos{\dfrac{\pi}{8}}. Therefore:

r0=eamp+amiq8a8=eampeamiq8a8.r1=eimz18a8. \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}

Let us do some working to simplify this last line. imz1=ame7πi8=amcosπ8+amisinπ8=amq+amipimz_1 = ame^{\dfrac{7\pi i}{8}} = -am\cos{\dfrac{\pi}{8}}+ami\sin{\dfrac{\pi}{8}} = -amq+amip. Therefore:

r1=eamq+amip8a8=eamqeamip8a8.r2=eimz28a8. \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}

Working on this last line. imz2=ame9πi8=amqamipimz_2 = ame^{\dfrac{9\pi i}{8}} = -amq-amip. Therefore:

r2=eamqamip8a8=eamqeamip8a8.r3=eimz38a8. \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}

Working on this last line. imz3=ame11πi8=ampamiqimz_3 = ame^{\dfrac{11\pi i}{8}} = -amp-amiq. Therefore:

r3=eampamiq8a8=eampeamiq8a8. \begin{aligned} r_3 &= -\dfrac{e^{-amp-amiq}}{8a^8} \\ &= -\dfrac{e^{-amp}e^{-amiq}}{8a^8}. \end{aligned}

Thus the sum of the residues of the poles within the region enclosed by C is:

n=03rn=18a8(eampeamiq+eamqeamip+eamqeamip+eampeamiq)=18a8(eamp(eamiq+eamiq)+eamq(eamip+eamip))=18a8(2eampcosamq+2eamqcosamp)=14a8(eampcosamq+eamqcosamp). \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}

Contour integral evaluation

Therefore our contour integral is:

Ceimzz(z8+a8)dz=2πi14a8(eampcosamq+eamqcosamp)=πi2a8(eampcosamq+eamqcosamp). \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}

Subintegral evaluation

Splitting our contour integral into subintegrals, we obtain:

Ceimzz(z8+a8)dz=AB+BD+DE+EA \begin{aligned} \oint_C \dfrac{e^{imz}}{z(z^8+a^8)} dz &= \int_{AB} + \int_{BD} + \int_{DE} + \int_{EA} \end{aligned}

where each integral on the right-hand side of course shares the same integrand as that on the left.

Along AB, z=xz=x, dz=dxdz=dx and xx goes from R-R to ϵ-\epsilon.

AB=Rϵeimxx(x8+a8)dx. \begin{aligned} \int_{AB} = \int_{-R}^{-\epsilon} \dfrac{e^{imx}}{x(x^8+a^8)} dx. \end{aligned}

When ϵ0\epsilon \rightarrow 0 and RR\rightarrow \infty:

AB=0eimxx(x8+a8)dx. \int_{AB} = \int_{-\infty}^0 \dfrac{e^{imx}}{x(x^8+a^8)} dx.

Along BD, z=ϵeiθz=\epsilon e^{i\theta}, dz=iϵeiθdθdz=i\epsilon e^{i\theta} d\theta and θ\theta goes from π\pi to 00.

BD=π0eimϵeiθϵeiθ(ϵ8e8iθ+a8)iϵeiθdθ=i0πeimϵeiθϵ8e8iθ+a8dθ. \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}

When ϵ0\epsilon \rightarrow 0:

BD=i0πdθa8=πia8. \begin{aligned} \int_{BD} &= -i \int_0^{\pi} \dfrac{d\theta}{a^8} \\ &= -\dfrac{\pi i}{a^8}. \end{aligned}

Along DE: z=xz=x, dz=dxdz=dx and xx goes from ϵ\epsilon to RR.

DE=ϵReimxx(x8+a8)dx. \begin{aligned} \int_{DE} &= \int_{\epsilon}^R \dfrac{e^{imx}}{x(x^8+a^8)} dx. \end{aligned}

When ϵ0\epsilon \rightarrow 0 and RR\rightarrow \infty:

DE=0eimxx(x8+a8)dx. \int_{DE} = \int_{0}^{\infty} \dfrac{e^{imx}}{x(x^8+a^8)} dx.

Finally, along EA, z=Reiθz=Re^{i\theta}, dz=iReiθdz=iRe^{i\theta} and θ\theta goes from 00 to pipi. Therefore:

EA=0πeimReiθReiθ(R8e8iθ+a8)iReiθdθ=i0πeimRcosθmRsinθR8e8iθ+a8dθ=i0πeimRcosθemRsinθR8e8iθ+a8dθ. \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}

For 0θπ0\leq \theta \leq \pi (our bounds of integration), emRsinθ1e^{-mR\sin{\theta}} \leq 1, and eimRcosθ=1|e^{imR\cos{\theta}}| = 1 provided m0m \geq 0. Therefore an upper bound on our integral is:

EAi0πdθR8e8iθ+a8 \begin{aligned} \int_{EA} \leq i \int_0^{\pi} \dfrac{d\theta}{R^8 e^{8i\theta}+a^8} \end{aligned}

The magnitude of that denominator is:

R8e8iθ+a8=(R8cos8θ+a8)2+R16sin28θ=R16+a16+2R8a8cos8θR16+a162R8a8=R8a8EAi0πdθR8a8=πiR8a8. \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}

As RR\rightarrow \infty, it is obvious therefore that EA0\displaystyle \int_{EA} \rightarrow 0.

Putting the pieces together

Therefore the sum of our subintegrals (which is equal to our contour integral) is:

Ceimzz(z8+a8)dz=eimxx(x8+a8)dxπia8 \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}

Replacing the left-hand side with what the residue theorem tells us this contour integral equals yields:

πi2a8(eampcosamq+eamqcosamp)=eimxx(x8+a8)dxπia8eimxx(x8+a8)dx=πi2a8(2eampcosamqeamqcosamp)cosmx+isinmxx(x8+a8)dx=πi2a8(2eampcosamqeamqcosamp)cosmxx(x8+a8)dx+isinmxx(x8+a8)dx=πi2a8(2eampcosamqeamqcosamp). \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}

Equating the imaginary part of this expression yields:

sinmxx(x8+a8)dx=π2a8(2eampcosamqeamqcosamp)=π2a8(2eamsinπ8cos(amcosπ8)eamcosπ8cos(amsinπ8)). \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}

Which is what we were required to determine.