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$$\int^{\infty}_{-\infty}\frac{\sin(ax)}{x(x^2+1)}dx=\pi(1-e^{-a}), \ a\ge0$$

I tried to solve but came up with $\pi(2-e^{-a}) $. Could you tell me where did I do the mistake?

if $x=z$ then $dz=dx$

$$\int_\gamma \frac{e^{iaz}}{z(z^2+1)}\quad and\quad z(z+i)(z-i)=0\quad \rightarrow z=0,z=\pm i$$

for $ z=0$ $$Res(f,0)=\lim_{z\to 0}\frac{z.e^{iaz}}{z(z^2+1)}=1$$

for $z=1$

$$Res(f,i)=\lim_{z\to i}\frac{(z-i).e^{iaz}}{z(z+i)(z-i)}=\lim_{z\to i}\frac{e^{iaz}}{z(z+i)}=\frac{e^{-a}}{-2}$$

$$\int_\gamma \frac{e^{iaz}}{z(z^2+1)}=\int_{-R}^{R}\frac{e^{iax}}{x(x^2+1)}dx+\int_\gamma \frac{e^{iaz}}{z(z^2+1)}=\pi i(1-\frac{e^{-a}}{-2})$$

$$\int_{-R}^{R}\frac{e^{iax}}{x(x^2+1)}=\int^{R}_{-R}\frac{\cos(ax)}{x(x^2+1)}dx+i\int_{-R}^R \frac{\sin(ax)}{x(x^2+1)}=i(2\pi - 2\pi \frac{e^{-a}}{2})$$

$$\rightarrow \int_{-R}^R\frac{\sin(ax)}{x(x^2+1)}=2\pi -2\pi \frac{e^{-a}}{2}=\pi(2-e^{-a}) $$

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    $\begingroup$ Your problem is that the pole in $z=0$ belongs to $[-R,R]$ so you cannot use the residue theorem here. $\endgroup$ – Tom-Tom Jan 14 '14 at 10:27
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    $\begingroup$ An approach is to make the pole at zero disappear, considering $f(z)=(\mathrm e^{\mathrm iaz}-1)/(z(z^2+1))$. The imaginary part still yields the desired integral. Completing the loop by upper halfcircles, one is left with a unique pole at $\mathrm i$, with residue $\frac12(1-\mathrm e^{-a})$. This assumes the contribution of the upper halfcircle part disappears when $R\to+\infty$, which happens when $a\geqslant0$. $\endgroup$ – Did Jan 14 '14 at 10:32
  • $\begingroup$ @lyme Plot the function $\dfrac{sin(ax)}{x(x^2-1)}$ for $a=1$. You can guess from this picture that the integral must be positive. But the formula you have $\pi(1-e^a)$ gives a negative value. This suggest that there is something wrong with your formula. $\endgroup$ – Braindead Jan 15 '14 at 0:26
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The integral depends on the value of $a$, which I assume is a real number.

Assuming that $a\ge 0$, You can use this kind of contour to integrate $f(z) = \dfrac{e^{iaz}}{z(z^2+1)}$. The point is that you have to go around the singularity at $z=0$. (An alternative is to follow Did's suggestion).

Using this contour, you obtain:

$$2\pi i Res(f,i) = \left(\int_R - \int_\epsilon\right) + \left(\int_r^R f(x) dx + \int_{-R}^{-r} f(x) dx\right),$$

$\int_R$ and $\int_\epsilon$ both represent integral of $f$ over the respective semicircles. A quick computation shows that $R\to \infty$ makes $\int_R\to 0$ and $\epsilon \to 0$ makes $\int_\epsilon\to i\pi$.

Therefore,

$$2\pi i Res(f,i) = - i\pi + \int_{-\infty}^\infty f(x) dx$$

Computing the residue and solving for the integral gives:

$$ \int \dfrac{\sin(ax)}{x(x^2+1)} dx = \pi (1-e^{-a})$$

For $a<0$, you can use a similar contour, but it has to be upside down from the one used above in order for the big $R$ contour to vanish as $R\to \infty$. The answer comes out to be $\pi (e^a -1)$.

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I will assume $a > 0$. Call the integral $I(a)$. Differentiate w.r.t. $a$, $$ I'(a) = \int_{-\infty}^{\infty} \frac{\cos(a x)}{1+x^2} \ \mathrm{d}x = 2\int_{0}^{\infty}\frac{\cos(a x)}{1+x^2} \ \mathrm{d}x,$$ then note that $$\mathcal{L}(I'(a)) = 2\int_{0}^{\infty}\int_{0}^{\infty} \frac{\cos(a x)}{1+x^2}\exp(-as) \ \mathrm{d}a \ \mathrm{d}x,$$ $$\mathcal{L}(I'(a)) = 2\int_{0}^{\infty} \frac{s}{(x^2+s^2)(1+x^2)} \ \mathrm{d}x.$$ Using partial fractions and integrating, $$\mathcal{L}(I'(a)) = \frac{\pi}{s+1},$$ taking the inverse transform gives $$ I'(a) = \pi \mathrm{e}^{-a}, $$ integrate to obtain $$ I(a) = -\pi \mathrm{e}^{-a} + c, $$ and using $I(0) = 0$, we see that $c=\pi$ and $$ I(a) = \pi(1-\mathrm{e}^{-a}). $$

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  • $\begingroup$ Watch out: the derivation of the formula for $I'(a)$ was for $a > 0$, so observing $I(0) = 0$ doesn't immediately tell you the value of $c$ unless you verify that $I(a) \rightarrow 0$ as $a \rightarrow 0^+$. To show that, write $I(a) = 2\int_0^\infty \sin(ax)/(x(x^2+1))dx$ and then use the inequality $|\sin(ax)| \leq ax$ for $x \geq 0$ to see that $|I(a)| \leq 2a\int_0^\infty dx/(x^2+1)$ for $a > 0$. Now let $a \rightarrow 0^+$ in the equation $I(a) = c - \pi{e}^{-a}$ to conclude that $0 = c - \pi$, so $c = \pi$. $\endgroup$ – KCd Jan 25 '14 at 23:33
  • $\begingroup$ That is very true. Thanks for calling me out on that! I did consider that I needed a limiting approach, but I generally write answers without any paper next to me, and I was too lazy :P. Thanks for correcting the error. $\endgroup$ – Bennett Gardiner Jan 25 '14 at 23:41
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}{\sin\pars{ax} \over x\pars{x^{2} + 1}}\,\dd x =\pi\sgn\pars{a}\pars{1 - \expo{-\verts{a}}}:\ {\large ?}.\qquad a \in {\mathbb R}}$

\begin{align} &\color{#00f}{\large\int_{-\infty}^{\infty}{\sin\pars{ax} \over x\pars{x^{2} + 1}} \,\dd x} =a\int_{-\infty}^{\infty}{1 \over x^{2} + 1}\, \color{#c00000}{\sin\pars{ax} \over ax}\,\dd x \\[3mm]&=a\int_{-\infty}^{\infty}{1 \over x^{2} + 1}\, \pars{\color{#c00000}{\half\int_{-1}^{1}\expo{\ic kax}\,\dd k}}\,\dd x =\half\,a\int_{-1}^{1}\pars{% \int_{-\infty}^{\infty}{\expo{\ic\verts{ka}x} \over x^{2} + 1}\,\dd x}\,\dd k \\[3mm]&=\half\,a\int_{-1}^{1}\pars{% 2\pi\ic\,{\expo{\ic\verts{ka}\ic} \over 2\ic}}\,\dd k =\half\,\pi a\int_{-1}^{1}\expo{-\verts{ka}}\,\dd k =\pi a\int_{0}^{1}\expo{-\verts{a}k}\,\dd k \\[3mm]&=\pi a\,{\expo{-\verts{a}} - 1 \over -\verts{a}} =\color{#00f}{\large\pi\sgn\pars{a}\pars{1 - \expo{-\verts{a}}}} \end{align}

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