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As far as I understand, I have to calculate integrals

$$\int_{0}^{\infty} \frac{1}{\sqrt{x}}\cos \omega x \operatorname{d}\!x$$


$$\int_{0}^{\infty} \frac{1}{\sqrt{x}}\sin \omega x \operatorname{d}\!x$$

Am I right? If yes, could you please help me to integrate those? And if no, could you please explain me.

EDIT: Knowledge of basic principles and definitions only is supposed to be used.

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The first one I am not sure, is this integral convergent? It creates issues at 0. The second one can be perhaps be done with Laplace transform. You do need the gamma function to deal with that squareroot. – imranfat May 27 '13 at 19:02
This is the task from the textbook and all previous tasks were relevant to current knowledge, which doesn't particularly include Laplace transform etc., so I hope, there is a simpler way (Anyway, it's a definite integral, so - easier way may still exist. – tkroman May 27 '13 at 19:05
up vote 2 down vote accepted


$$\int_0^{\infty} dx \, x^{-1/2} e^{i \omega x}$$

Substitute $x=u^2$, $dx=2 u du$ and get

$$2 \int_0^{\infty} du \, e^{i \omega u^2}$$

The integral is convergent, and may be proven so using Cauchy's theorem. Consider

$$\oint_C dz \, e^{i \omega z^2}$$

where $C$ is a wedge of angle $\pi/4$ in the first quadrant and radius $R$. This integral over the closed contour is zero, and at the same time is

$$\int_0^R dx \, e^{i \omega x^2} + i R \int_0^{\pi/4} d\phi \, e^{i \phi} e^{-\omega R^2 \sin{2 \phi}} e^{i \omega R^2 \cos{2 \phi}} + e^{i \pi/4} \int_R^0 dt \, e^{-\omega t^2} = 0$$

The second integral, because $\sin{2 \phi} \ge 4 \phi/\pi$, has a magnitude bounded by $\pi/(4 \omega R)$, which vanishes as $R \to \infty$. Therefore

$$\int_0^{\infty} dx \, e^{i \omega x^2} = e^{i \pi/4} \int_0^{\infty} dt \, e^{-\omega t^2} = e^{i \pi/4} \sqrt{\frac{\pi}{\omega}}$$


$$\int_0^{\infty} dx \, x^{-1/2} e^{i \omega x} = (1+i)\sqrt{\frac{2 \pi}{\omega}}$$

The Fourier cosine and sine transforms follow from taking the real and imaginary parts of the above. Note the dependence on $\omega^{-1/2}$ times some scale factor.

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What a nice solution, thank you very much. Exactly within boundaries of my current knowledge but enough tricky for me to miss the bus. Thank you! – tkroman May 27 '13 at 19:28
@cdshines: you're welcome. Please remember to accept an answer that is useful to you. – Ron Gordon May 27 '13 at 19:31
That's a clever solution indeed. I am jst wondering why the differential is immediately behind the integral sign and not behind the integrated expression? It looks kind a odd... – imranfat May 27 '13 at 19:33
You could also integrate $f(z) = z^{-\frac{1}{2}} e^{i \omega z}$ around a wedge of angle $\frac{\pi}{2}$ that is indented at the origin. – Random Variable May 27 '13 at 19:33
@imranfat: it is common in physics and especially optics which is my background. You are hardly the first to ask about this (even today!). – Ron Gordon May 27 '13 at 19:35

First, use the change of variables $ y=\omega \,x $ $$I=\int_{0}^{\infty} \frac{1}{\sqrt{x}}\cos \omega x\, dx = \frac{1}{\sqrt{\omega} }\int_{0}^{\infty} \frac{1}{\sqrt{y}}\cos y\,dy\,.$$

Then, use the Mellin transform method (using the tables)

$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x) dx .$$

Now, the Mellin transform of $\cos(y)$ is $$ \Gamma \left( s \right) \cos \left( \frac{\pi}{2} \,s \right) .$$

Then subs $s=\frac{1}{2}$, since $s-1=-\frac{1}{2}$, gives

$$ I = \frac{1}{\sqrt{\omega}}\Gamma \left( \frac{1}{2} \right) \cos \left( \frac{\pi}{4} \right) . $$

You can do the same for the other one.

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This is really great, but, as I said, this task is given just after definitions of FT and sine/cosine transforms, so, I suppose, no Mellin transforms are accepted. – tkroman May 27 '13 at 19:12
First line: $\int f(x) {\bf dx}$ and +1 very nice twist... – draks ... May 27 '13 at 19:13
@cdshines: You did not mention this in your post, but it is ok. It is a good technique to learn. – Mhenni Benghorbal May 27 '13 at 19:15
@draks...: thanks for the comment. I really appreciate it. – Mhenni Benghorbal May 27 '13 at 19:16
I've edited my post to avoid further confusions, thank you. – tkroman May 27 '13 at 19:19

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