Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

How to prove this result?

share|improve this question
    
Check this technique. –  Mhenni Benghorbal Mar 26 at 19:58
add comment

4 Answers 4

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 $b=0$.

share|improve this answer
    
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 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 $0$. –  Sami Ben Romdhane Mar 26 at 19:58
1  
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 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 at 20:15
    
Try to differentiate $\arctan\frac b a$ with respect to $a$ and you'll see why we have this negative sign. –  Sami Ben Romdhane Mar 26 at 20:18
show 1 more comment

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

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}

share|improve this answer
add comment

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

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

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

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

Then

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

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

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

share|improve this answer
add comment

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

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.