Evaluating the integral $\int_0^{\infty} dk\frac{\sin(kr)}{k(k^2 + m^2)}$

Question

I have been trying to solve Poisson's Equation, given by $$\nabla^2\Phi(\vec{r}) = -\frac{\rho(\vec{r})}{\epsilon_0}$$ for the given charge density $$\rho(\vec{r}) = \frac{\rho_0}{m}\frac{e^{-mr}}{r}$$ for positive parameters $\rho_0$ and $m$, subject to the boundary condition that $\Phi(\vec{r})\rightarrow 0$ as $r\rightarrow\infty$. To do this, I did a Fourier Transform of the initial equation, which I then solved for the Fourier Transform of $\Phi(\vec{r})$, $\hat{\Phi}(\vec{k})$ to obtain: $$\hat{\Phi}(\vec{k}) = \frac{4\pi\rho_0}{m\epsilon_0}\frac{1}{m^2 + k^2}$$ When applying the Inverse Fourier Transform, the integral in question arises as the last integral to calculate - $\Phi(\vec{r})$ is given by $$\Phi(\vec{r}) = \frac{\rho_0}{\pi\epsilon_0 mr}\int_{0}^{\infty}dk\frac{\sin(kr)}{k(k^2 + m^2)}$$ I have entered this integral into Wolfram Alpha, which gives me the result $$\Phi(\vec{r}) = \frac{\rho_0}{\epsilon_0 m^3 r}(1 - e^{-mr})$$ This agrees with the result one obtains when calculating $\Phi(\vec{r})$ directly using the formula obtained by using the Green Function of the $\nabla^2$-Operator, so I know that it is correct.

However, I'd still be interested in how one would go about evaluating this integral, because none of the integration techniques that I know seem to work at all.

Any help would be much appreciated.

Answer 1

According to Plancherel's formula \begin{equation*} \int_{-\infty}^{\infty}\overline{f(x)}g(x)\, dx = \dfrac{1}{2\pi}\int_{-\infty}^{\infty}\overline{F(\xi)}G(\xi)\, d\xi \end{equation*} where $f,g\in L_{2}(\mathbb{R})$ and $F(\xi) = \int_{-\infty}^{\infty}f(x)e^{-i\xi x}\, dx$ (Fourier transform).

If $f(x) = \dfrac{\sin(xr)}{x}$ then $F(\xi) = \pi(H(\xi+r)-H(\xi-r))$, where $H$ is the Heaviside step function. Futhermore, if $g(x) = \dfrac{1}{x^2+m^2}$ then $G(\xi) = \dfrac{\pi}{m}e^{-|m\xi|}$.

Thus \begin{equation*} \int_{-\infty}^{\infty}\dfrac{\sin(xr)}{x(x^2+m^2)}\, dx = \dfrac{1}{2\pi}\int_{-r}^{r}\dfrac{\pi^2}{m}e^{-|m\xi|}\, d\xi = \dfrac{\pi}{m}\int_{0}^{r}e^{-m\xi}\, d\xi = \dfrac{\pi}{m^2}(1-e^{-mr}). \end{equation*} Finally \begin{equation*} \Phi = \dfrac{\rho_{0}}{2\epsilon_{0}m^{3}r}(1-e^{-mr}) \end{equation*} with an extra 2 factor.

Answer 2

Have you tried with Residues Calculus?

Your integral is an improper integral with trigonometric function. First of all notice that your function is even, hence you may write

$$\frac{1}{2}\int_{-\infty}^{+\infty} \frac{\sin(kr)}{k(k^2 + m^2)}\ \text{d}k$$

Actually, for those kinds of integral, what we are really doing is to evaluate the so called Principal Value, that is

$$\lim_{R\to\infty} \frac{1}{2}\int_{-R}^{+R} \frac{\sin(kr)}{k(k^2 + m^2)}\ \text{d}k$$

Indeed called the Cauchy principal Value. But let's leave this apart for the moment.

Now you can evaluate it with Residues Theorem. First of all, the sine at the numerator is nothing but the imaginary part of the complex exponential:

$$\sin(kr) = \Im e^{ikr}$$

So we write

$$\frac{1}{2}\Im \int_{-\infty}^{+\infty} \frac{e^{ikr}}{k(k^2 + m^2)}\ \text{d}k$$

Now the integral has three poles at $k = 0$, $k = +im$ and $k = -im$. Indeed you may rewrite the denominator in a clearer way to see it:

$$\frac{1}{2}\Im \int_{-\infty}^{+\infty} \frac{e^{ikr}}{k(k+im)(k-im)}\ \text{d}k$$

Hence we have by Residues Calculus

$$\frac{1}{2}\Im\left\{2\pi i \sum_i\ \text{Res}[f(k), k_i]\right\}$$

Where $k_i$ are the poles.

The first pole is at $k = 0$ so we have

$$2\pi i\left(\lim_{k\to 0} k\cdot \frac{e^{ikr}}{k(k+im)(k-im)}\right) = \frac{2\pi i}{im(-im)} = \frac{2\pi i}{m^2}$$

For the other two poles $k = im$ and $k = -im$ we only consider the poles who lie in the upper half plane, hence $k = im$:

$$2\pi i\left(\lim_{k\to im} (k-im)\cdot \frac{e^{ikr}}{k(k-im)(k+im)}\right) = 2\pi i \frac{e^{i(im)r}}{(im)(im + im)} = 2\pi i\frac{e^{-mr}}{2i^2m^2} = -\pi i \frac{e^{-mr}}{m^2}$$

Collecting all the residues together:

$$\pi i\left(\frac{2}{m^2} - \frac{e^{-mr}}{m^2}\right)$$

Hence

$$\frac{1}{2}\Im\left\{2\pi i \sum_i\ \text{Res}[f(k), k_i]\right\} = \frac{\pi}{2m^2}\left(2 - e^{-mr}\right)$$

Using this into your result:

$$\Phi(\vec{r}) = \frac{\rho_0}{\pi\epsilon_0 mr}\int_{0}^{\infty}dk\frac{\sin(kr)}{k(k^2 + m^2)} = \frac{\rho_0}{\pi\epsilon_0 mr}\cdot \frac{\pi}{2m^2}\left(2 - e^{-mr}\right)$$

$$\Phi(\vec{r}) = \frac{\rho_0}{2\epsilon_0 r m^3}\left(2 - e^{-mr}\right)$$

Which is ... not your solution lol.

Something went wrong, somewhere. I will send this answer the same, just to give you some hints and help about how to compute the integral, even if there is that $2$ factor more.

Learn Residues Calculus. It's mandatory.

Answer 3

Hint. You are in fact interested in evaluating $$ I(a)=\int_{0}^{\infty}\frac{\sin(ax)}{x(x^2 + 1)}\:dx,\qquad a>0.\tag1 $$ Let's consider the Laplace transform of $\displaystyle \int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx$. We have $$ \begin{aligned} \mathcal{L}\left(\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx\right)(s) & = \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}e^{-as}\:da\:dx \\&= \int_{0}^{\infty}\frac{s}{(x^2+1)(s^2+x^2)}\;{dx} \\&= \frac{\pi}{2(s+1)} \end{aligned} $$giving $$ \int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx=\mathcal{L}^{-1}\left(\frac{\pi}{2(s+1)}\right) =\frac{\pi}{2}e^{-a},\qquad a>0, \tag2 $$ then by differentiating $(2)$ with respect to $a$, one gets $$ \int_{0}^{\infty}\frac{x\sin(ax)}{x^2+1}\:dx=\frac{\pi}{2}e^{-a},\qquad a>0. \tag3 $$ By partial fraction decomposition we have $$ \frac{1}{x(x^2+1)}=\frac{1}{x}-\frac{x}{x^2+1} $$ and $$ \begin{align} \int_{0}^{\infty}\frac{\sin(ax)}{x(x^2 + 1)}\:dx&=\int_{0}^{\infty}\frac{\sin(ax)}{x}\:dx-\int_{0}^{\infty}\frac{x\sin(ax)}{x^2+1}\:dx \\\\&=\frac{\pi}{2}-\frac{\pi}{2}e^{-a},\qquad a>0, \end{align} $$ where we have used the classic result $$ \int_{0}^{\infty}\frac{\sin(ax)}{x}\:dx\stackrel{u=ax}{=}\int_{0}^{\infty}\frac{\sin(u)}{u}\:du=\frac{\pi}{2},\qquad a>0. $$ Finally,

$$ \int_{0}^{\infty}\frac{\sin(ax)}{x(x^2 + 1)}\:dx=\frac{\pi}{2}\left(1-e^{-a}\right),\qquad a>0. $$

You then get your initial integral with the change of variable, $u=bx$ $$ \int_{0}^{\infty}\frac{\sin(a u)}{u(u^2 + b^2)}\:du=\frac1{b^2}\int_{0}^{\infty}\frac{\sin(abx)}{x(x^2 + 1)}\:dx=\frac{\pi}{2b^2}(1 - e^{-ab}) $$ as announced.