# Trig Fresnel Integral

$$\int_{0}^{\infty }\sin(x^{2})dx$$

I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you.

I have not done complex analysis (only real analysis as I am a high school student) so how can I evaluate it using elementary functions (without complex analysis)?

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I'm not sure, would $\dfrac{1}{x^2} \times (-cos(x^2))$ be correct? –  Niklas R May 1 '13 at 8:38
This is called a Fresnel integral. The standard method to evaluate it uses complex analysis, but I am sure there are more elementary methods available if you Google for them. –  Potato May 1 '13 at 8:38
@Aryabhata: I am not sure it is an exact duplicate. Unless I have missed something, the answer to the question being asked here cannot just be read off... –  user1729 May 1 '13 at 9:32
@Aryabhata It can be read off the very first sentence so long as you know you need to use complex analysis. So there is a leap. But I suppose that is covered by Potato's comment. –  user1729 May 1 '13 at 9:38
I edited my question. –  please delete me May 1 '13 at 12:01

$\displaystyle \int_{0}^{\infty}\sin(x^{2})dx$

Let $\displaystyle u=x^{2}, \;\ x=\sqrt{u}, \;\ dx=\frac{1}{2}u^{-1/2}du$

$\displaystyle\frac{1}{2}\int_{0}^{\infty}\frac{\sin(u)}{\sqrt{u}}du$

Now, the Gamma function comes in handy:

$\displaystyle\frac{1}{2\Gamma(a)}\int_{0}^{\infty}z^{a-1}e^{-uz}\sin(u)dz$

But, since $a=1/2$, we have:

$\displaystyle\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-uz}\sin(u)}{\sqrt{z}}dudz$

$\displaystyle\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{1}{\sqrt{z}(1+z^{2})}dz$

Let $t=\sqrt{z}$:

$\displaystyle\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{1}{t^{4}+1}dt$

This is a rather famous integral and can be found here and there. I leave it's evaluation to the reader.

But, it evaluates to $\displaystyle \frac{1}{\sqrt{\pi}}\cdot \frac{\pi\sqrt{2}}{4}=\frac{\sqrt{2\pi}}{4}$, and so does the Fresnel in question.

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I'm not completely following how you get equation 3 from equation 2. Can you help me out? –  Bitrex May 3 '13 at 23:04
It comes from the Gamma function. You can use this as a parameter in the integral. $\frac{1}{u^{a}}=\frac{1}{\Gamma(a)}\int_{0}^{\infty}z^{a-1}e^{-uz}dz$ –  Cody May 4 '13 at 11:24
I follow now. Thanks! –  Bitrex May 4 '13 at 15:25
An nice integration, if you're interested, is to derive $\int_{0}^{\infty}\cos(x^{a})dx, \;\ a>1$. You can do it with Laplace transforms instead of the classic residue method. –  Cody May 5 '13 at 13:33

The integral does not have an elementary antiderivative. If you want to avoid complex analysis, try this method.

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