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Can one prove that $$ \int_0^\infty x^{d-4}\sin x\, dx = \cos \frac{\pi d}{2} \Gamma(d-3),\text{ for }2<\Re(d)<4? $$ I would prefer using the methods of contour integration.

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    $\begingroup$ set $d=3$ then the integral evaluates to $\frac{\pi}{2}$. But the right hand-side is zero. see this for the integral in this case: math.stackexchange.com/questions/5248/… $\endgroup$ – Math-fun Jan 15 '16 at 17:45
  • $\begingroup$ @Math-fun You are right; I am terribly sorry but a typo must have propagated via copy-paste: it's $\Gamma(d-3)$. $\endgroup$ – Brightsun Jan 15 '16 at 18:18
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Consider the contour $\Gamma$ given by the quarter-circle of radius $M$, centred at the origin and lying in the first quadrant, with a quarter-circle indent of radius $\varepsilon$ at the origin. The function $$ f(\zeta) = e^{i\zeta}\zeta^{d-4}, \text{ for } 3<\Re(d)<4, $$ is holomorphic within such a contour and therefore $$ \oint_\Gamma f(\zeta) d\zeta=0. $$ Writing explicitly the integration along the four parts of the contour, we have: $$ 0=\int_\varepsilon^M e^{ix} x^{d-4}dx + i M^{d-3}\int_0^{\pi/2} e^{iMe^{i\varphi}}e^{i(d-3)\varphi}d\varphi\\ - i^{d-3}\int_\varepsilon^M y^{d-4}e^{-y}dy -i\varepsilon^{d-3}\int_0^{\pi/2}e^{i\varepsilon e^{i\varphi}}e^{i(d-3)\varphi}d\varphi. $$ The integral on the arc of radius $M$ is controlled by the following argument: letting $\Re(d)=\xi$ and $\Im(d)=\eta$ $$ \left| i M^{d-3}\int_0^{\pi/2} e^{iMe^{i\varphi}}e^{i(d-3)\varphi}d\varphi \right|\le M^{\xi-3}\int_0^{\pi/2}e^{-M\sin\varphi-\eta \varphi}d\varphi; $$ now, as long as $0\le\varphi\le\pi/2$, we have $\sin\varphi\ge2\varphi/\pi$, so that $$ M^{\xi-3}\int_0^{\pi/2}e^{-M\sin\varphi-\eta \varphi}d\varphi \le M^{\xi-3}\int_0^{\pi/2}e^{-(2M/\pi+\eta)\varphi}d\varphi\\ =\frac{\pi M^{\xi-3}}{2M+\pi \eta}\left( 1- e^{-M-\eta\pi/2}\right)=\frac{\pi M^{\xi-4}}{2+\pi \eta M^{-1}} \left( 1- e^{-M-\eta\pi/2}\right), $$ which tends to zero as $M\to\infty$, since we have $\xi<4$. For the same reason, $$ \left| i\varepsilon^{d-3}\int_0^{\pi/2}e^{i\varepsilon e^{i\varphi}}e^{i(d-3)\varphi} \right| \le \frac{\pi \varepsilon^{\xi-3}}{2\varepsilon+\pi \eta}\left( 1- e^{-\varepsilon-\eta\pi/2}\right); $$ this behaves as $$\frac{\varepsilon^{\xi-3}}{\eta} (1-e^{-\eta\pi/2})\text{ if }\eta\neq 0$$ and as $$\pi\varepsilon^{\xi-3}(1-e^{-\varepsilon})(2\varepsilon)^{-1}\approx\pi \varepsilon^{\xi-3}/2\text{ for }\eta=0$$ and hence goes to zero as $\varepsilon\to0$ for any $\eta$, since $\xi>3$. Thus, for $3<\Re(d)<4$, $$ \int_0^\infty e^{ix} x^{d-4} dx = i^{d-3}\int_0^{\infty}y^{d-4}e^{-y}dy; $$ since $i^{d-3}=ie^{i\pi d/2}$, and $$ \int_0^{\infty}y^{d-4}e^{-y}dy = \Gamma(d-3), $$ equating separately real and imaginary parts we obtain $$ \int_0^\infty \sin x\, x^{d-4} dx = \cos(\pi d/2) \Gamma(d-3), $$ which establishes the result for $3<\Re(d)<4$, and $$ \int_0^\infty \cos x\, x^{d-4} dx = -\sin(\pi d/2) \Gamma(d-3). $$ Now, we integrate by parts in the second equation: $$ -\sin(\pi d/2) \Gamma(d-3)= \sin x\, x^{d-4}\Big|_0^\infty - (d-4) \int_0^\infty x^{d-5} \sin x\, dx; $$ since, $3<\xi<4$, the boundary contribution vanishes, so $$ \int_0^\infty x^{d-5} \sin x\, dx = \sin \frac{\pi d}{2}\frac{\Gamma(d-3)}{d-4} $$ and relabelling $d=D+1$, using $\sin (\pi (D+1)/2) = \cos(\pi D/2)$ and $\Gamma(d-3)=(d-4)\Gamma(d-4)$, we have $$ \int_0^\infty x^{D-4} \sin x\, dx = \cos \frac{\pi D}{2} \Gamma(D-3), $$ establishing the result for $2<\Re(D)<3$.

The identity for $D=3$ reduces to show (letting $d=3+\epsilon$) $$ \int_0^{\infty}\frac{\sin x}{x}dx = \lim_{\epsilon\to0}\cos \frac{3\pi + \epsilon \pi}{2}\Gamma(\epsilon), $$ but $$ \cos \frac{3\pi + \epsilon \pi}{2}\Gamma(\epsilon) = \sin(\epsilon \pi/2) \Gamma(\epsilon) \approx \frac{\pi \epsilon}{2}\epsilon^{-1}, $$ so that one needs only prove $$ \int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}, $$ which again follows by elementary contour integration.

Since the right-hand side of the identity has no other singularity for $\Re(d)=3$, this establishes the formula in the whole strip $2<\Re(d)<4$.

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