# $\int_0^\infty x e^{-\mathrm i x\cos(\varphi)}\mathrm dx=-\frac{1}{\cos (\varphi )^2}$ is that correct?

Good day.
This integral looks very simple, yet I don't know how to start. $$\int_0^\infty x e^{-\mathrm i x\cos(\varphi)}\mathrm dx$$ I know that if the lower integration limit was $-\infty$ it would be a derivative of a Dirac delta function $2\pi \mathrm i\;\delta'(\cos(\varphi))$. But it isn't. So I guess it should somehow relate to Dirac delta.
So how to cope with this one. Can I use a Laplace transform? Or I thought about using Heaviside function under the integral, but it didn't work. :(

UPDATE
I've got a result using Heaviside function but don't know whether it is correct. $$I=\int_0^\infty x e^{-\mathrm i x\cos(\varphi)}\mathrm dx=\int_{-\infty}^\infty x e^{-\mathrm i x\cos(\varphi)}u(x)\mathrm dx$$ where $u(x)$ is the Heaviside function.
So using its' limiting reperesentation: $$I=\int_{-\infty}^\infty x e^{-\mathrm i x\cos(\varphi)}u(x)\mathrm dx=\frac{1}{2\pi \mathrm i}\lim_{\epsilon\to0^+}\int_{-\infty}^\infty x e^{-\mathrm i x\cos(\varphi)}\int_{-\infty}^\infty\frac{e^{\mathrm i x\tau}}{\tau-\mathrm i \epsilon}\mathrm d\tau \; \mathrm dx$$

$$I=\frac{1}{2\pi \mathrm i}\lim_{\epsilon\to0^+}\int_{-\infty}^\infty \frac{\mathrm d\tau}{\tau-\mathrm i \epsilon}\int_{-\infty}^\infty xe^{-\mathrm i x(\cos(\varphi)-\tau)}\mathrm dx=\frac{1}{2\pi \mathrm i}\lim_{\epsilon\to0^+}\int_{-\infty}^\infty \frac{\mathrm d\tau}{\tau-\mathrm i \epsilon}(2\pi\mathrm i \delta'(\cos(\varphi)-\tau))$$ Then I use the property $\int_{-\infty}^\infty \delta'(x)\varphi(x)\,\mathrm dx = -\int_{-\infty}^\infty \delta(x)\varphi'(x)\,\mathrm dx$ $$I=\lim_{\epsilon\to0^+}\frac{1}{(\epsilon +i (\cos (\varphi )))^2}=-\frac{1}{\cos (\varphi)^2}$$ Does that make any sense?

By parts:

$$u=x\;\;,\;\;u'=1\\v'=e^{-ix\cos\phi}\;,\;\;v=-\frac1{i\cos\phi}e^{-ix\cos\phi}\;\;,$$

thus

$$\int\limits_0^\infty xe^{-ix\cos\phi}dx=\left.\frac{xi}{\cos\phi}e^{-ix\cos\phi}\right|_0^\infty-\frac i{\cos\phi}\int\limits_0^\infty e^{-ix\cos\phi}dx=$$

$$=\left.\frac1{\cos^2\phi}e^{-ix\cos\phi}\right|_0^\infty=-\frac1{\cos^2\phi}$$

Note: observe that the above is correct only if $\,\cos\phi\in\Bbb C\setminus\Bbb R\;$ and, in fact, only if $\;\text{Im}\,(\cos\phi)<0\;$ , otherwise the integral diverges...!

• and for $\phi\in\Bbb R$ no hope? – user91450 Sep 26 '13 at 11:39
• I can't see it...that $\;i\;$ there screws things up because if $\;\cos\phi\in\Bbb R\;$ then that exponential is an element of the canonical unit circle... – DonAntonio Sep 26 '13 at 11:46
• Physicist don't mind evaluating divergent integrals! – GEdgar Sep 26 '13 at 12:06
• @GEdgar well, this is from my research work and it comes from statistics (in physics). But still, how can I cope with is?! – user91450 Sep 26 '13 at 12:23
• @user91450 :$$\lim_{b\to\infty}\int\limits_0^b xe^{-ix\cos\phi}dx$$ But $$\lim_{b\to\infty}\left|\frac{bi}{\cos\phi e^{ib\cos\phi}}\right|=\lim_{b\to\infty}\frac b{\cos\phi}=\pm \infty\;\;(\text{depending on the sign of}\;\cos\phi)$$ So in this case I think even CPV doesn't exist finitely... – DonAntonio Sep 26 '13 at 14:08