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$$\int_0^{\infty}x^{-1}e^{-ax}\sin (bx) \;\mathrm dx = \arctan \frac{b}{a}$$

How to prove this result?

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Check this technique. – Mhenni Benghorbal Mar 26 '14 at 19:58

Let $$F(a)=\int_0^{\infty}x^{-1}e^{-ax}\sin (bx) \;\mathrm dx $$ then we can prove using Leibniz theorem: differentiate under the sign $\int$ that:

$$F'(a)=-\int_0^{\infty}e^{-ax}\sin (bx) \;\mathrm dx=-\operatorname{Im} \int_0^{\infty}e^{(-a+ib)x} \;\mathrm dx=\operatorname{Im}\frac{1}{-a+ib}=-\frac b{a^2+b^2}$$ so $$F(a)=-\int \frac b{a^2+b^2}da=\arctan\frac b a+C$$ Notice that $C=0$ since the integral is zero for $b=0$.

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I trying to follow your steps because I like that approach, but when anti deriving that e-power, I need to plug in infinity for $x$ in the term $e^{(-a+bi)x}$ and that is supposed to be zero, as you are getting the $Im$ term for $x=0$. Could you explain how this e-power vanishes when x is put infinity? – imranfat Mar 26 '14 at 19:55
Take the absolute value you find $e^{-ax}$ and its limit is clearly $0$. Notice that if the absolute value converges to $0$ the complex number converges also to $0$. – user63181 Mar 26 '14 at 19:58
O yes, that makes sense. The modulus goes to zero and the argument keeps going between 0 and 360 and so the whole complex number becomes zero, I see it now when working it out... – imranfat Mar 26 '14 at 20:12
What about that negative in front of that integral in the last sentence? That shouldn't be there because there are 2 negatives that cancel which is why you get a positive arctan – imranfat Mar 26 '14 at 20:15
Try to differentiate $\arctan\frac b a$ with respect to $a$ and you'll see why we have this negative sign. – user63181 Mar 26 '14 at 20:18

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}x^{-1}\expo{-ax}\sin\pars{bx}\,\dd x = \arctan\pars{b \over a}:\ {\large ?}}$

Assumming $\ds{a > 0}$: \begin{align} &\color{#00f}{\large\int_{0}^{\infty}x^{-1}\expo{-ax}\sin\pars{bx}\,\dd x}= \sgn\pars{b}\int_{0}^{\infty}\exp\pars{-\,{a \over \verts{b}}\,x}\, {\sin\pars{x} \over x}\,\dd x \\[3mm]&= \sgn\pars{b}\int_{0}^{\infty}\exp\pars{-\,{a \over \verts{b}}\,x}\, \pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}\,\dd x \\[3mm]&= \half\sgn\pars{b}\int_{-1}^{1}\braces{\int_{0}^{\infty} \exp\pars{\bracks{-\,{a \over \verts{b}} + \ic k}x}\,\dd x}\,\dd k = \half\sgn\pars{b}\int_{-1}^{1}{1 \over a/\verts{b} - \ic k}\,\dd k \\[3mm]&= \sgn\pars{b}\int_{0}^{1}{a/\verts{b} \over \pars{a/b}^{2} + k^{2}}\,\dd k = \sgn\pars{b}\sgn\pars{a}\int_{0}^{\verts{b/a}}{1 \over k^{2} + 1}\,\dd k \\[3mm]&=\sgn\pars{b \over a}\arctan\pars{\verts{b \over a}} =\color{#00f}{\large\arctan\pars{b \over a}} \end{align}

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For $\text{Re}(s) >0$ and $a,b >0$,

$$ \int_{0}^{\infty} x^{s-1} e^{-ax} \sin(bx) \ dx = - \text{Im} \int_{0}^{\infty} x^{s-1} e^{-(a+ib)x} \ dx$$

$$ =- \text{Im} \ \mathcal{L}_{t} [x^{s-1}](a+ib) = - \text{Im} \frac{\Gamma (s)}{(a+ib)^{s}}$$

$$ = - \text{Im} \ \Gamma(s) \frac{e^{-is \arctan (\frac{b}{a})}}{(a^{2}+b^{2})^{s/2}} = \frac{\Gamma (s)}{(a^{2}+b^{2})^{s/2}} \sin \left(s \arctan (\frac{b}{a}) \right)$$


$$ \int_{0}^{\infty}\frac{e^{-ax} \sin (bx)}{x} \ dx = \lim_{s \to 0^{+}}\frac{\Gamma (s)}{(a^{2}+b^{2})^{s/2}} \sin \left(s \arctan (\frac{b}{a}) \right) $$

$$ = \lim_{s \to 0^{+}} \Big( \frac{1}{s} + \mathcal{O}(s) \Big) \sin \left(s \arctan (\frac{b}{a}) \right)$$

$$ = \lim_{s \to 0^{+}} \frac{\sin (s \arctan \frac{b}{a})}{s} = \arctan \left(\frac{b}{a} \right)$$

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Use Parseval's Theorem:

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$

where $f$ and $F$ are Fourier transform pairs, as are $g$ and $G$. We identify

$$f(x) = e^{-a x} \theta(x) \implies F(k) = \frac1{a-i k}$$ $$g(x) = \frac{\sin{b x}}{x} \implies G(k) = \begin{cases}\pi & |k| \le b \\ 0 & |k| \gt b \end{cases}$$

where $\theta(x)$ is the Heaviside step function ($0$ when $x \lt 0$, $1$ when $x \gt 0$).Then the integral is

$$\frac12 \int_{-b}^b \frac{dk}{a-i k} = \frac{i}{2} \log{\left (\frac{1-i b/a}{1+i b/a} \right )} = \arctan{\frac{b}{a}}$$

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the question has $\int_0^\infty$. Do you think that might be OP's typo? WA is unable to evaluate. – Sabyasachi Mar 26 '14 at 19:18
No, the transform is right, I just forgot to include the Heaviside. – Ron Gordon Mar 26 '14 at 19:19

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