Integral of delta dirac function

Question

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?

Answer 1

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 : $$\tag{4}I=\frac {\pi}{b^{\,5}\;e^{\large{\frac ab}}}$$

Answer 2

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.