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For any $A>0$,$|\int_0^A{\sin(x)}|\le2$ and ${\frac{1}{\sqrt{x}}}$ is decreasing to $0$ as $x$ tend to $+\infty$.By Dirichlet's test,the integration $\int_0^{+\infty}{\frac{\sin(x)}{\sqrt{x}}}dx$ makes sense.

Moreover,I know the method of computing $\int_0^{+\infty}{\frac{\sin(x)}{x}}dx$ which is that considering the integration $$I(\beta)=\int_0^{+\infty}f(x,\beta)dx=\int_0^{+\infty}e^{-\beta{x}}{\frac{\sin(x)}{x}}dx$$ and $$I^{'}(\beta)=\int_0^{+\infty}f_\beta(x,\beta)dx=-\int_0^{+\infty}e^{-\beta{x}}{\sin(x)}dx.$$ We can compute $\int_0^{+\infty}e^{-\beta{x}}{\sin(x)}dx$ easily thourgh integral by parts.If we use the same method,we will obtain$$I^{'}(\beta)=-\int_0^{+\infty}e^{-\beta{\sqrt{x}}}{\sin(x)}dx.$$ Integral by parts is out of use since it will produce many $\sqrt{x}$.Can you tell me how to integrate this integration and more general case $\int_0^{+\infty}{\frac{\sin(x)}{x^a}}dx$,where $0<a<1$.Thank you in advance!

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marked as duplicate by Daniel Fischer, Sujaan Kunalan, dfeuer, Lukas Geyer, Praphulla Koushik Nov 23 '13 at 17:44

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2 Answers 2

up vote 2 down vote accepted

Let $t^2=x:$

$$\int_0^{\infty}\frac{\sin x}{\sqrt{x}}\,dx=\int_0^{\infty} 2\sin t^2\,dt=\sqrt{\frac{\pi}{2}}$$

Where the latter integral is the Fresnel integral.

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How poor my searching skill is! Thank you very much! –  F.G Nov 23 '13 at 15:51

You can find here Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\pi/2}$ solutions.

Note: I don't have enough reputation for comment.

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