Find $ \int_{0}^{\infty} e^{ix} \sin(x) \frac{e^{-3x}}{x} dx$

Question

The second contribution in the Born approximation for the Yukawa potential in scattering theory leads to the following integral (for some given ratio of parameters): \begin{align} \int_{0}^{\infty} e^{ix} \sin(x) \frac{e^{-3x}}{x} dx \end{align} The solution is said to be (wolframalpha) $\tan^{-1}(\frac{3}{10} + \frac{i}{10})$. How is this derived? Is there a way to evaluate the numerical value of the integral (e.g. via the residue theorem?)

Derivation of the integral: Starting point was an integral equation for the wave function of the scattering problem described by the Schroedinger equation:

\begin{align} \phi_{\vec{k}}(\vec{r}) = \frac{1}{(2\pi)^{\frac{3}{2}}} e^{i \vec{k} \cdot \vec{r}} - \frac{m}{2\pi \hbar^2} \int^{}_{} d^3 r' \frac{e^{ik|\vec{r} -\vec{r}'|}}{|\vec{r} - \vec{r}'|} V(\vec{r}') \phi_{\vec{k}}(\vec{r}') \end{align}

Simply inserting plane waves in the RHS leads to what is called the Born approximation: \begin{align} \phi_{\vec{k}}(\vec{r}) = \frac{1}{(2\pi)^{\frac{3}{2}}} e^{i \vec{k} \cdot \vec{r}} - \frac{m}{2\pi \hbar^2} \int^{}_{} d^3 r' \frac{e^{ik|\vec{r} -\vec{r}'|}}{|\vec{r} - \vec{r}'|} V(\vec{r}') \bigg(\frac{1}{(2\pi)^{\frac{3}{2}}}e^{i \vec{k} \vec{r}'} \bigg) \end{align}

In class, we derived the following expression for the scattering amplitude, and furthermore used the Born approximation to show that it is the fourier transformed of the potential.

\begin{align} f_{\vec{k}} (\theta, \phi) = - \frac{\sqrt{2\pi}m}{ \hbar^2} \int^{}_{} d^3 r' e^{-i \vec{k}'(k, \theta, \phi) \vec{r}'} V(\vec{r}') \phi_{\vec{k}}(\vec{r}') \approx -\frac{m}{2\pi \hbar^2} \int^{}_{} d^3 r' e^{+i(\vec{k} - \vec{k}'(k,\theta, \phi)) \vec{r}'} V(\vec{r}') \end{align}

In the lecture notes a formula is given to evaluate when the Born approximation is good. It starts from the above formula, assumes a symmetric potential with respect to rotation, goes over to spherical coordinates and most importantly assumes $\vec{r} =0$ for a potential centered at 0. The last assumption is justified as the influence of the potential should be the highest at this choice. This leads to the criterium \begin{align} \frac{2m}{\hbar^2 k} \bigg|\int^{\infty}_{0} dr' e^{ikr'} V(r') \sin(kr')\bigg| \ll 1 \end{align} The Yukawa potential reads $V(r) = A \frac{e^{-\lambda r}}{r}$. The task was to numerically (I am just curious for an analytical perspective) evaluate for which $A$ the Born approximation is good with $\frac{k}{\lambda} = 3$. The substitution $x = kr'$ leads to my initial question.

Answer 1

$I(a) = \int_0^\infty \frac {e^{-ax}\sin x}{x} \ dx$

We need to find $I(3-i)$

Differentiation under the integral sign.

$I'(a) = \int_0^\infty -e^{-ax}\sin x \ dx\\ I'(a) = \frac {e^{-ax}(a\sin x + \cos x)}{a^2+ 1}|_0^\infty\\ I'(a) = -\frac {1}{a^2+ 1}\\ I(\infty) - I(3-i) = \int_{3-i}^\infty -\frac {1}{a^2+ 1}\ da =\arctan (3-i) - \frac{\pi}{2}\\ I(3-i) = \frac{\pi}{2}-\arctan (3-i)$

How to reconcile with the W-A answer

$\frac{\pi}{2}-\arctan (3-i) = \arctan (\frac 3{10} + \frac {i}{10})$

Take tan of both sides:

$\tan (\frac{\pi}{2}-\arctan (3-i)) = \frac 3{10} + \frac {i}{10}\\ \tan (\frac {\pi}{2} - x) = \frac {1}{\tan x}\\ \frac {1}{3-i} = \frac 3{10} + \frac {i}{10}$

Answer 2

The Direct Method

Denote the given integral as $I$, itrodice $a=3-i$ and use $1/x=\int_{0}^{1} t^{x-1} dt$ $$I=\int_{0}^{1} \frac{dt}{t} \int_{0}^{\infty} e^{-ax} \sin x~ t^x ~dx=\int_{0}^{1} \frac{dt}{t} \int_{0}^{\infty} e^{-x(\ln (t^{-1}e^a))} \sin x ~dx$$ $$\Rightarrow I=\int_{0}^{1} \frac{dt}{t} \int_{0}^{\infty} \exp[-x(\ln (t^{-1}e^a))]~ (2i)^{-1} [e^{ix}-e^{-ix}] ~dx.$$ $$\Rightarrow I= (2i)^{-1} \int_{0}^{1} \frac{dt}{t} \left( \frac{1}{\ln (t^{-1}e^a)-i}- \frac{1}{\ln (t^{-1}e^a)+i}\right)=\int_{0}^{1} \frac{dt/t}{\ln^2(t^{-1}e^a) +1}.$$ Letting $\ln(t^{-1} e^a) =u \ \Rightarrow t=e^{a-u} \Rightarrow dt=-e^a e^{-u} du.$ We get $$I=-\int_{\infty}^{a} \frac{du}{1+u^2}=\pi/2-\tan^{-1} a= \tan^{-1} \frac{1}{a}=\tan^{-1}\frac{1}{3-i}=\tan^{-1} \frac{3+i}{10} .$$