Integral of delta dirac function I try to calculate the following integral:
\begin{equation}
\int^{+\infty}_{-\infty} \frac{x^4 \exp{(ixa)}}{1+x^2b^2} \mathrm{d}x
\end{equation}
where $a,b$ are real positive numbers.
This integral does not converge and I hope that it can be represented by some kind of Delta dirac function. But I don't know how to do it. Any suggestions?
 A: As you said, the integral does not converge, it does not have a rigorous meaning. If you really want to write something other than $\infty$ on the right-hand side, though, you may proceed as follows. 
Formally, one can define
$$
\int_{-\infty}^{+\infty}\frac{x^4e^{ixa}}{1+x^2b^2}dx \equiv \left(\frac{\partial }{\partial a} \right)^4\int_{-\infty}^{+\infty}\frac{e^{ixa}}{1+x^2b^2}dx;
$$
now, the integral appearing on the right-hand side is convergent (in fact, absolutely convergent) and we may compute it using complex analysis and contour integration: use a half-circle $C_M$ of radius $M$ in the upper-half plane centred at the origin. By the residue theorem, since the only singularity of denominator lying within $C_M$ is a pole at $z_+=i/b$ (thanks to the fact that $b$ is positive), we have 
$$
\oint_{C_M}\frac{e^{i\zeta a}}{1+\zeta^2b^2}d\zeta = 
i2\pi \text{Res}\frac{e^{i\zeta a}}{1+\zeta^2b^2}\Big|_{z=i/b}=\frac{i\pi}{b^2(i/b)}e^{-a/b}=\frac{\pi}{b}e^{-a/b}.
$$ 
Now, dividing the integration contour into the real segment going from $-M$ to $M$ and the big arc of radius $M$, we have
$$
\int_{-M}^{+M}\frac{e^{ixa}}{1+x^2b^2}dx + 
\int_0^\pi \frac{e^{iM e^{i\varphi} a}}{1+M^2e^{i2\varphi}b^2}iMe^{i\varphi}d\varphi
=\frac{\pi}{b}e^{-a/b}.
$$
The integral on the arc vanishes as $M\to\infty$ since, exploiting the fact that $a>0$,
$$
\left|
\int_0^\pi \frac{e^{iM e^{i\varphi} a}}{1+M^2e^{i2\varphi}b^2}iMe^{i\varphi}d\varphi
\right|
\le \int_0^\pi \frac{e^{-Ma\sin\phi}}{M^2\left|M^{-2}+e^{i2\varphi}b^2\right|}Md\varphi\le
\int_0^\pi \frac{1}{M^2\left|M^{-2}-b^2\right|}Md\varphi = \frac{\pi}{M|b^2-M^{-2}|}.
$$
Hence
$$
\int_{-\infty}^{+\infty}\frac{e^{ixa}}{1+x^2 b^2}dx = \frac{\pi}{b}e^{-a/b},\text{ for }a,b>0.
$$
By our definition, differentiating four times with respect to $a$,
$$
\int_{-\infty}^{+\infty}\frac{x^4e^{ixa}}{1+x^2b^2}dx  = \frac{\pi}{b^5}e^{-a/b}.
$$
As final remark, if we let $a\equiv |\xi|$, for $\xi\in\mathbb R$, we have, for any test function $\psi$ of fast decrease:
$$
\int_{-\infty}^{+\infty} \frac{\pi}{b^5}e^{-|\xi|/b}\psi(\xi) d\xi = 
\int_{-\infty}^{+\infty} \frac{\pi}{b}e^{-|\xi|/b}\psi^{(4)}(\xi) d\xi \\
=\int_{-\infty}^{+\infty} \pi e^{-|s|}\psi^{(4)}(b s) ds \xrightarrow[b\to0^+]{}\psi^{(4)}(0)\int_{-\infty}^{+\infty} \pi e^{-|s|}ds = 2\pi \psi^{(4)}(0),
$$
where $\psi^{(4)}$ means the fourth derivative of $\psi$.
This means that, with our definition,
$$
\lim_{b\to0^+}\int_{-\infty}^{+\infty}\frac{x^4e^{ix|\xi|}}{1+x^2b^2}dx =2\pi \delta^{(4)}(\xi)
$$
in the sense of tempered distributions.
A: Let's ignore $b$ first (a simple change of variable allows to 'reinstall' it) then :
$$\tag{1}I(a):=\int^{+\infty}_{-\infty} \frac{\exp{(ixa)}}{1+x^2} \mathrm{d}x=\mathcal{F}\frac 1{1+x^2}\left(a\right)=\frac {\pi}{e^a}$$
so that :
$$\tag{2}\int^{+\infty}_{-\infty} \frac{\exp{(ixa)}}{1+x^2b^2} \mathrm{d}x=\frac{I(a/b)}b=\frac {\pi}{b\;e^{\large{\frac ab}}}$$
Should you now choose to ignore the divergence of the integral then you will get :
\begin{align}
I&:=P.V.\int^{+\infty}_{-\infty} \frac{x^4\exp{(ixa)}}{1+x^2b^2} \mathrm{d}x\\
I&=P.V.\int^{+\infty}_{-\infty} \frac{(ix)^4\exp{(ixa)}}{1+x^2b^2} \mathrm{d}x\\
\tag{3}I&=\left(\frac d{da}\right)^4\frac {\pi}{b\;e^{\large{\frac ab}}}\\
\end{align}
that should be :
$$\tag{4}I=\frac {\pi}{b^{\,5}\;e^{\large{\frac ab}}}$$
A: In terms of distributions, splitting off the polynomial part, we get
$$\mathcal F\left[ \frac {x^4} {1 + b^2x^2} \right] =
\left( \frac {x^4} {1 + b^2x^2}, e^{i a x} \right) = \\
\pi |b|^{-5} e^{-|a/b|} -2 \pi b^{-2} \delta''(a) - 2 \pi b^{-4} \delta(a),$$
which indeed can be identified with the ordinary function $\pi b^{-5} e^{-a/b}$ for positive $a,b$.
