Does the integral $\int_0^t \frac{x}{(b^2x^2+1)^2} \sin(ax)dx$ converge? I tried to compute this integral in Python, but it shows no results.
I used the integration by parts and it led to the following integral
$\displaystyle \int_0^t \frac{\cos(ax)}{b^2x^2+1} \,dx$, which is still complicated.
 A: The integral exists, as commented, because the integrand is continuous on $[0,t]$.  However, it is not an elementary function, so your basic methods of integration won't evaluate it for you.
Maple says (if $a>0,b>0$) the value of the integral is
$$
\frac {1}{2\,b} \left( -i\cosh \left( {\frac {a}{b}} \right) {\it Ci}
 \left( -{\frac {a \left( -bt+i \right) }{b}} \right) +{\it Si}
 \left( {\frac {a \left( -bt+i \right) }{b}} \right) \sinh \left( {
\frac {a}{b}} \right)\\ +i\cosh \left( {\frac {a}{b}} \right) {\it Ci}
 \left( {\frac {a \left( bt+i \right) }{b}} \right) +\pi\,\cosh
 \left( {\frac {a}{b}} \right) -{\it Si} \left( {\frac {a \left( bt+i
 \right) }{b}} \right) \sinh \left( {\frac {a}{b}} \right)  \right)
$$
Here, $Si$ and $Ci$ are the
Sine Integral and Cosine Integral functions.
The limit as $t \to +\infty$ is
$$
-{\frac {\pi}{2\,b} \left( \sinh \left( {\frac {a}{b}} \right) -\cosh
 \left( {\frac {a}{b}} \right)  \right) }
=
{\frac {\pi}{2\,b}{{\rm e}^{-{a/b}}}}
$$
A: I don't see why do we have to evaluate the integral to prove convergence.
We just write $|\int_{0}^{t}\dfrac{xsinaxdx}{(b^{2}x^{2}+1)^{2}}|$ $\leq$
$\int_{0}^{t}|\dfrac{xsinaxdx}{(b^{2}x^{2}+1)^{2}}|$ $\leq$$\int_{0}^{t}\dfrac{x|sinax|dx}{(b^{2}x^{2}+1)^{2}}$
$\leq$$\int_{0}^{t}\dfrac{xdx}{(b^{2}x^{2}+1)^{2}}$=$\dfrac{1}{2}\int_{0}^{t}\dfrac{d(x^{2})}{(b^{2}x^{2}+1)^{2}}$=
$\dfrac{1}{2b^{2}}\int_{0}^{t}\dfrac{d(b^{2}x^{2}+1)}{(b^{2}x^{2}+1)^{2}}$
=$\dfrac{1}{2b^{2}}$$[-\dfrac{1}{(b^{2}x^{2}+1)}]_{0}^{t}$=$-\dfrac{1}{2b^{2}}(\dfrac{1}{b^{2}t^{2}+1}-1)$
which converges to $\dfrac{1}{2b^{2}}$ as $t\to +\infty$.
Now consider the case $b=0$. Then the integral for $a=0$ clearly converges since it is zero.
If $a\neq\,0$ then the integral is equal to:$\dfrac{sinat-atcosat}{a^{2}}$
which does not converge.
