<s>[INST] What is the Fourier Sine Transform of $\frac{1}{x(a^2+x^2)}$? (Please check my solution) My textbook says that the answer is $\frac{\pi}{2a^2}(1-e^{-as})$ while my answer is $\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$. Here is my solution:
\begin{align}
\hat{f_s} & = \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{1}{t(a^2+t^2)} \sin{st}dt \\
 & = \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{t}{a^2}[\frac{1}{t^2} -\frac{1}{(a^2+t^2)}] \sin{st}dt \\
 & = \frac{1}{a^2}\sqrt{\frac{2}{\pi}} \int_0^\infty [\frac{\sin{st}}{t} -\frac{t}{(a^2+t^2)} \sin{st}] dt \\
 & = \frac{1}{a^2}\sqrt{\frac{2}{\pi}} \left( \frac{\pi}{2} - \int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt \right) 
\end{align}
Let this integral be $y$. Therefore,
\begin{align}
y & = \int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt  \\
 & = \int_0^\infty \frac{1}{t}[1 -\frac{a^2}{(a^2+t^2)}] \sin{st}dt \\
 & = \int_0^\infty [\frac{\sin{st}}{t} -\frac{a^2}{t(a^2+t^2)} \sin{st}] dt \\
 & = \left( \frac{\pi}{2} - \int_0^\infty \frac{a^2}{t(a^2+t^2)} \sin{st} dt \right) 
\end{align}
Differentiating $y$ w.r.t. $s$ twice, we get
\begin{align}
y'' & = a^2\int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt  \\
 & = a^2 y \\
\end{align}
This is a second order differential equation, whose solution is 
$$y= C_1e^{-as}+C_2e^{as}$$
Put $s=0$ in $y$ and $y'$, we get $C_1 = \frac{\pi}{4} = - C_2$. Thus, we get $$y = \frac{\pi}{4}(e^{-as}-e^{as})$$ or
$$\hat{f_s}=\frac{1}{a^2}\sqrt{\frac{2}{\pi}}\left(\frac{\pi}{2} - \frac{\pi}{4}(e^{-as}-e^{as})\right)$$ which is nothing but $$\hat{f_s}=\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$$
 [/INST] When doing the change of variable $u = sx$, it is implicitly assumed that $s \neq 0$ (otherwise, the integral evaluates to $0$). This means that the middle expression of $y$ and the subsequent chain of differentiations (hence, also, the obtained expression for $y''$) are only valid for $s \neq 0$. To obtain a formula for all $s$, we must differentiate:
$$y = \int_0^{\infty} \frac{t}{t^2 + a^2} \sin(st)dt$$
Now there is a problem if we want to differentiate under the integral sign, because the integrand is not integrable in the sense of Lebesgue
Edit
The added content is elaboration on Olivier Oloa's comment on the original post. 
As suggested, we can differentiate:
$$I(s) = \int_0^{\infty} \frac{1}{t(t^2 + a^2)} \sin(st)dt$$
to get:
$$I'(s) = \int_0^{\infty} \frac{\cos(st)}{a^2 + t^2}dt$$
(because the integrand is Lebesgue integrable and its partial derivative wrt $s$ is uniformly bounded, wrt $t$, by an integrable function)
One way to obtain $I'(s)$ is by using the Fourier transform of $g(t) = e^{-a|t|}$. It is given by:
$$\mathcal{F} (g)(t) =  \sqrt{\frac{2}{\pi}}\frac{a}{a^2 + t^2}$$
Using the inversion formula, we get:
$$\frac1{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \sqrt{\frac{2}{\pi}}\frac{a}{a^2 + t^2} e^{-it s}dt = e^{-a|s|}$$
Hence:
$$\int_{-\infty}^{\infty} \frac{e^{-it s}}{a^2 + t^2} dt = \frac{\pi}a e^{-a|s|}$$
Assuming $s \ge 0$, we finally get:
$$I'(s) = \frac{\pi}{2a} e^{-as}$$
So 
$$I(s) = \frac{\pi}{2a^2} (C - e^{-as})$$
As $I(0) = 0$, we get $C  = 1$.
</s>