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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?

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  • $\begingroup$ Could you please provide us with some information as to the values $a$ and $b$ are allowed to attain? $\endgroup$ – Brightsun Jan 13 '16 at 12:26
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    $\begingroup$ I added this information in my post. $\endgroup$ – user2863620 Jan 13 '16 at 12:28
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    $\begingroup$ I dont believe there is any hope of expressing this as some kind of Dirac delta function. The "singularities" are at $-\infty$ and $\infty$, and no matter what you do with $a,b$, the integrad will not go to 0 "outside $\infty$". $\endgroup$ – user159517 Jan 13 '16 at 12:49
  • $\begingroup$ to isolate the divergent part, i would try to add a small imaginary part to $a\rightarrow a\pm i \epsilon $ and see what happens in the end $\endgroup$ – tired Jan 13 '16 at 13:51
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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.

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    $\begingroup$ +1: neat observation concerning the Dirac distribution. Cheers, $\endgroup$ – Raymond Manzoni Jan 13 '16 at 13:51
  • $\begingroup$ very elegant derivation! Thanks! $\endgroup$ – user2863620 Jan 13 '16 at 14:13
  • $\begingroup$ @user2863620 my pleasure. If you like an answer to a question of yours, here on MathSE, you are also allowed to "accept" it, by clicking on the check sign at the top left of it, in case you think it is the best one suiting your request. :) Bye! $\endgroup$ – Brightsun Jan 13 '16 at 14:31
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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 : $$\tag{4}I=\frac {\pi}{b^{\,5}\;e^{\large{\frac ab}}}$$

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  • $\begingroup$ I have a question: is the $P.V.$ of the OP's integral, really well defined? As far as I can see, the Cauchy principal value only takes care of setting $\lim_{\alpha\to+\infty}\int_{-\alpha}^{+\alpha}x^4(1+x^2b^2)^{-1}i\sin(xa)$ to zero, by parity, but still has the diverging contribution coming from $\lim_{\alpha\to+\infty}\int_{-\alpha}^{+\alpha}x^4(1+x^2b^2)^{-1}\cos(xa)$. In other words, I can't see how $\cos(ax)$ could provide cancellation of oscillations of amplitude $x^2$ at infinity... $\endgroup$ – Brightsun Jan 13 '16 at 13:15
  • $\begingroup$ Yes but you ignore my previous sentence "Should you now choose to ignore the divergence of the integral" : the integral is indeed divergent (this is merely a 'regularization' and the P.V. cancels indeed only the imaginary part). $\endgroup$ – Raymond Manzoni Jan 13 '16 at 13:23
  • $\begingroup$ Ok, I'm sorry, I was just confused by the use of the P.V. notation in the third equation: it seemed as though the right-hand side was meant to be well-defined. Thank you very much for your clarification! $\endgroup$ – Brightsun Jan 13 '16 at 13:25
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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 $a,b$.

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