Difficult integral

**Gina88** · July 21st 2009, 11:05 PM

How can I calculate the integral of sin(x^2) from zero to infinity? Do I need to use the residue theorem?
I need a detailed answer, because i'm very new in this subject.
Thanks for any kind of help

**DeMath** · July 21st 2009, 11:34 PM

Originally Posted by Gina88

How can I calculate the integral of sin(x^2) from zero to infinity? Do I need to use the residue theorem?
I need a detailed answer, because i'm very new in this subject.
Thanks for any kind of help

View Wikipedia Fresnel integral

**Gina88** · July 22nd 2009, 03:21 AM

I'm sorry, but I didn't understand.
I know I can write sin(x^2) like this:x^2 - x^6/3! + x^10/5! - x^14/7!...
And then what?

**mr fantastic** · July 22nd 2009, 05:36 AM

Originally Posted by Gina88

I'm sorry, but I didn't understand.
I know I can write sin(x^2) like this:x^2 - x^6/3! + x^10/5! - x^14/7!...
And then what?

Integrate $e^{iz^2}$ over the contour C shown in the link you were given. Note that by Cauchy's theorem $\int_C e^{iz^2} \, dz = 0$ .

**NonCommAlg** · July 22nd 2009, 07:20 AM

start with the well-known and easy to prove intgeral $\int_0^{\infty}e^{-x^2} \ dx = \frac{\sqrt{\pi}}{2}.$ let $x^2=st,$ where we choose $s$ to be the variable. then we'll get $\frac{1}{\sqrt{t}}=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{s}}e^{-st} \ ds. \ \ \ \ \ \ (1)$

also it's easy to see that $\int_0^{\infty} e^{-ax} \sin x \ dx = \frac{1}{a^2+1}. \ \ \ \ \ \ (2)$

let $J=\int_0^{\infty} \frac{dx}{\sqrt{x}(x^2+1)}$ and $x=z^2.$ then $J=2 \int_0^{\infty} \frac{dz}{z^4 + 1}.$ now let $u=z - \frac{1}{z}, \ \ v = z+\frac{1}{z}.$ then: $2\int \frac{dz}{z^4 + 1}=\int \frac{du}{u^2 + 2} + \int \frac{dv}{v^2-2}=\frac{1}{\sqrt{2}} \tan^{-1}(u/\sqrt{2}) + \frac{1}{2\sqrt{2}} \ln \left|\frac{v-\sqrt{2}}{v+\sqrt{2}} \right|,$

which gives us: $J=\frac{\pi}{\sqrt{2}}. \ \ \ \ \ \ \ (3)$

now let $I=\int_0^{\infty} \sin x^2 \ dx$ and put $x^2=t.$ then by (1): $I=\frac{1}{2} \int_0^{\infty} \frac{\sin t}{\sqrt{t}} \ dt = \frac{1}{2 \sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{s}} \int_0^{\infty} e^{-st} \sin t \ dt \ ds.$ so by (2) and (3): $I=\frac{1}{2 \sqrt{\pi}} \int_0^{\infty}\frac{ds}{\sqrt{s}(s^2+1)}=\frac{1} {2\sqrt{\pi}} \cdot \frac{\pi}{\sqrt{2}}=\sqrt{\frac{\pi}{8}}.$

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