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how could i evaluate the following integral ?? $$\int_{-\infty}^\infty dt \frac{\exp(-iut)}{|at|^{1/2+ib}}$$

here $a$ and $b$ are positive real numbers.. how can i make this integral ? thanks.

$ |x| $ means the absolute value function

I think this integral is related to the Mellin transform $$\int_0^\infty dx \cos(ux)x^{s-1}$$ with $ s=1/2-ib $ but i don not know how to get this Mellin transform

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up vote 4 down vote accepted

Let's define $\displaystyle\ F(b):= \int_{-\infty}^\infty \frac{e^{-it}}{|t|^{1/2+ib}}\,dt\ = 2\int_0^\infty \frac{\cos(t)}{t^{1/2+ib}}\,dt$

Since $\displaystyle \int_0^\infty \frac{\cos(t)}{t^z}\,dt= \sin\left(\frac{\pi z}2\right)\Gamma(1-z)\ \ $ (for $\Re(z) \in (0,1)$)

(you may indeed find this integral in tables of Mellin transforms or in tables of Fourier cosine transform : Gradshteyn and Ryzhik (17.34.6) for example)

This seems to be a result of Euler who provided the integral (Whittaker and Watson 'A course of Modern Analysis' page 260 example 12, see too the Hankel integral of Gamma §12.22) : $\displaystyle \int_0^{\infty}\frac {\cos(ux)}{x^z}\,dx=\frac {\pi}{2 \Gamma(z)} u^{z-1}\sec\left(\frac {\pi z}2\right)\ $

This corresponds to the previous result using $\frac {\pi}{\Gamma(z)}=\Gamma(1-z)\sin(\pi z)\ $ and $\ \sin(\pi z)=2\sin\bigl(\frac{\pi}2 z\bigr)\cos\bigl(\frac{\pi}2 z\bigr)$.

so that $F(b)$ becomes : $$F(b)= 2\,\sin\left(\frac {\pi}2\left(\frac 12 +ib\right)\right)\Gamma\left(\frac 12 - ib\right)$$

A change of variable of $t\to ut$ and a division by $|a|^{1/2+ib}$ should give your solution.

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OK thanks :) , by the way where i could get some info about the Cosine integral evaluation :) as you have pointed ? thanks. – Jose Garcia Jun 10 '12 at 21:20

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