Complex analysis for real valued integrals.

Question

I'm asked to compute

$$\int_{-\infty}^{\infty}\frac{\sin(ax)}{x(1+x^4)}dx$$

To do this, I consider the function

$$f(z)=\frac{e^{aiz}}{z(1+z^4)}$$

and the contour composed by the segment on the real positive line going from $\epsilon$ to $R$, the half-upper circumference, the segment from $-R$ to $-\epsilon$ and the half-lower circumference around zero. When the radius of the small circumference tends to zero, the border of the contour contains a pole of the function, $0$ in particular. What should I do in this case?

Can you please show me how to compute this integral? Thanks in advance.

Answer 1

Define $$f(z)=\frac{e^{aiz}}{z(1+z^4)}$$ Let $C_R, C_\epsilon$ to be the half-upper circumferences with radius $R,\epsilon$, respectively and $C_1, C_2$ to the segments from $-R$ to $-\epsilon$, $\epsilon$ to $R$, respectively. For big $R$ and small $\epsilon$, inside $C_R$-$C_1$-$C_\epsilon^-$-$C_2$, $f(z)$ has two poles $$ z_{1,2}=\pm\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, z_{1,2}^4=-1. $$ So $$ \int_{C_R}f(z)dz+\int_{C_1}f(z)dz -\int_{C_\epsilon}f(z)dz+\int_{C_2}f(z)dz=2\pi i\bigg[\text{Re}(f,z_1)+\text{Re}(f,z_2)\bigg]. $$ Note \begin{eqnarray} \text{Re}(f,z_1)+\text{Re}(f,z_2)=\frac{e^{aiz_1}}{4z_1^4}+\frac{e^{aiz_2}}{4z_2^4}=-\frac{1}{2}e^{-\frac{\sqrt2}{2}a}\cos(\frac{\sqrt2}{2}a) \\ \int_{C_1}f(z)dz=\int_{-R}^{-\epsilon}f(x)dx, \int_{C_2}f(z)dz=\int^{R}_{\epsilon}f(x)dx. \end{eqnarray} Since $$ \bigg|\int_{C_R}f(z)dz\bigg|\le\int_{C_R}|f(z)||dz|\le\frac{e^{-aR\sin\theta}}{R(R^4-1)}\pi R=\frac{\pi e^{-aR\sin\theta}}{R^4-1} $$ and $$ \int_{C_\epsilon}f(z)dz=\pi i\text{Re}(f,0)=\pi i$$ letting $R\to\infty,\epsilon\to0$, one has $$ \int_{-\infty}^\infty f(x)dx-\pi i= -\pi ie^{-\frac{\sqrt2}{2}a}\cos(\frac{\sqrt2}{2}a) $$ Taking imaginary parts in both sides gives $$ \int_{-\infty}^{\infty}\frac{\sin(ax)}{x(x^4+1)}dx=\pi- \pi e^{-\frac{\sqrt2}{2}a}\cos(\frac{\sqrt2}{2}a) $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{p_{1} \equiv \expo{\pi\ic/4}}$ and $\ds{p_{2} \equiv \expo{3\pi\ic/4}}$ are the integrand poles in the complex plane upper half. Then, \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty} {\sin\pars{ax} \over x\pars{1 + x^{4}}}\,\dd x} = \on{sgn}\pars{a}\int_{-\infty}^{\infty} {\sin\pars{\verts{a}x} \over x\pars{1 + x^{4}}}\,\dd x \\[5mm] = &\ \on{sgn}\pars{a}\,\Im\int_{-\infty}^{\infty} {\expo{\ic\verts{a}x} - 1 \over x\pars{1 + x^{4}}} \,\dd x \\[5mm] = &\ \on{sgn}\pars{a}\,\Im\braces{% 2\pi\ic\sum_{n = 1}^{2} \on{Res}\pars{{\expo{\ic\verts{a}x} - 1 \over x\bracks{1 + x^{4}}}}_{\ x\ =\ p_{k}}\,\,} \\[5mm] = &\ 2\pi\on{sgn}\pars{a}\,\Re\bracks{% \sum_{n = 1}^{2} {\expo{\ic\verts{a}p_{k}} - 1 \over p_{k} \pars{4p_{k}^{3}}}} \\[5mm] = &\ -\,{\pi \over 2}\on{sgn}\pars{a}\,\Re \sum_{n = 1}^{2} \pars{\expo{\ic\verts{a}p_{k}} - 1} \\[5mm] = &\ {\pi \over 2}\on{sgn}\pars{a} \left[\vphantom{\Large A}2\right. \\[2mm] &\ \!\!\!\!\!\left.-\Re\pars{% \expo{\ic\verts{a}\root{2}/2 - \verts{a}\root{2}/2} + \expo{-\ic\verts{a}\root{2}/2 - \verts{a}\root{2}/2} \,\,\,}\right] \\[5mm] = &\ {\pi \over 2}\on{sgn}\pars{a}\ \times \\[2mm] &\ \bracks{ 2 - 2\exp\pars{-\,{\root{2}\verts{a} \over 2}} \cos\pars{\root{2}\verts{a} \over 2}} \\[5mm] = &\ \pi\on{sgn}\pars{a}\ \times \\[2mm] &\ \bracks{ 1 - \exp\pars{-\,{\root{2}\verts{a} \over 2}} \cos\pars{\root{2}a \over 2}} \end{align}