Evaluating the integral $\int_0^{\infty}\frac{x^3}{x^2+a^2}\,\mathrm{d}x$ 
If $\displaystyle\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x=\large\displaystyle\dfrac1{ka^6}$, then find the value of $\displaystyle\dfrac{k}{8}$.

I tried a lot but finally stuck at an intermediate form :
$$\begin{align}
&\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x, \text{with}\, {x^2=t},{2x~\mathrm{d}x=\mathrm{d}t}\\
&=\frac12\int_0^{\infty}\dfrac{(x^2)(2x)}{x^2+a^2}\,\mathrm{d}x=\frac12\int_0^{\infty}\dfrac{t}{t+a^2}\,\mathrm{d}t=\frac12\int_0^{\infty}\dfrac{t+a^2-a^2}{t+a^2}\,\mathrm{d}t\\
&=\frac12\left[\int_0^{\infty}\mathrm{d}t-\int_0^{\infty}\dfrac{a^2}{t+a^2}\,\mathrm{d}t\right]=\frac12\left[t|_0^{\infty}-a^2\ln(a^2+t)|_0^{\infty}\right]
\end{align}$$ 
 A: $$I={\displaystyle\int}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x$$
Substitute $u=x^2+a^2$ thus $\mathrm{d}x=\dfrac{1}{2x}\,\mathrm{d}u$
$$I=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{u-a^2}{u}\,\mathrm{d}u$$
$$I={\dfrac{1}{2}\displaystyle\int}\left(1-\dfrac{a^2}{u}\right)\mathrm{d}u$$
$$I=\dfrac{1}{2}{\displaystyle\int}1\,\mathrm{d}u-\dfrac{1}{2}\class{steps-node}{\cssId{steps-node-2}{a^2}}{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$$
$$I=\dfrac{u}{2}-\dfrac{a^2\ln\left(u\right)}{2}+c$$
$$I=\left(\dfrac{x^2+a^2}{2}-\dfrac{a^2\ln\left(x^2+a^2\right)}{2}\right)\biggr|_{x=0}^{\infty}$$
The integral is divergent.
A: $$I=\int_0^\infty\frac{x^3}{x^2+a^2}dx$$
firstly, let $u=x^2+a^2$ so $\frac{du}{dx}=2x \therefore dx=\frac{du}{2x}$
For $x=0$, $u=a^2$. For $x\to\infty$, $\lim_{x\to \infty}\left(x^2+a^2\right)\to\infty$
so our integral becomes:

$$I=\frac{1}{2}\int_{a^2}^\infty\frac{x^3}{x^2+a^2}.\frac{du}{x}=\frac{1}{2}\int_{a^2}^\infty\frac{x^2}{x^2+a^2}dx=\frac{1}{2}\int_{a^2}^\infty\frac{u-a^2}{u}du=\frac{1}{2}\int_{a^2}^\infty du-\frac{a^2}{2}\int_{a^2}^\infty\frac{1}{u}du$$

Both parts to this integral are divergent and so the integral cannot be calculated
A: The integral does not converge.
Let
$$ I=\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x.$$
By the Schwinger parametrization we have
$$ I=\int_0^\infty \mathrm{d}t\, \exp(-ta^2)\int_0^\infty \mathrm{d}x\, x^3 \exp\left(-tx^2\right).$$
The last integral can be calculated by the Feynman trick. Using this result, one gets
$$I=\int_0^\infty \mathrm{d}t\,\frac{ \sqrt{\pi}\exp(-ta^2)}{4 t^{3/2}}=\frac{\sqrt{\pi}a}{4}\Gamma \left(-1,0\right),$$
which is infinity since
$$ \Gamma\left(-1,0\right)\equiv\int_0^\infty \mathrm{d}t\, t^{-3/2}\exp(-t)=\tilde{\infty}.$$
