Is $\int\cos\left(\frac{\pi}{2}\cdot x^2\right)\,dx$ a known integral ?

I found on the net something called Fresnel integral, but we didn't learn it, and it also somehow related to Euler, and we didn't touch the Euler stuff, so maybe I made a mistake before it while calculating the double integral:

$$\iint_D\sin\left(\frac{\pi x}{2y}\right)\,dxdy$$ where $D=\left\{(x,y)\in\mathbb R^2: x\le y\le \sqrt[3]{x}\wedge y\ge\frac{\sqrt{2}}{2}\right\}$

So I wrote the $D$ as a simple to $x$:

$\begin{cases}\frac{\sqrt{2}}{2}\le y\le 1\\y^3\le x\le y\end{cases}$

and then I did the integral by $x$ and got that troublesome integral.

So can I solve this problem without Fresnel integral or maybe i have some mistake on the way?

  • 2
    $\begingroup$ $$\int_{0}^{+\infty}\sin(x^2)\,dx = \int_{0}^{+\infty}\cos(x^2)\,dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}$$ are Fresnel's integrals, not Fres$\color{red}{\text{h}}$nel's. Anyway, please improve formatting by using $\LaTeX$, your question is almost un-readable. $\endgroup$ – Jack D'Aurizio Jun 21 '15 at 16:59
  • $\begingroup$ One thing: while editing, I interpreted "sin(pi x/2y)" as $\sin\left(\frac{\pi x}{2y}\right)$. Strictly speaking, "sin(pi*x/2y)" = $\sin\left(\frac{\pi x}{2}y\right)$. Did I guess your mind correctly? $\endgroup$ – user228113 Jun 21 '15 at 17:10
  • $\begingroup$ see here mathworld.wolfram.com/FresnelIntegrals.html $\endgroup$ – Dr. Sonnhard Graubner Jun 21 '15 at 17:15
  • $\begingroup$ i ment as you corrected ) thanks. $\endgroup$ – Ilya.K. Jun 21 '15 at 17:17
  • $\begingroup$ I understood that it isnt Frehsnel. Can somebody try to solve this double integral ? I get an integral that looks like cos(x^2) and that integral i dont know how to solve ? What did i miss ? $\endgroup$ – Ilya.K. Jun 21 '15 at 17:21

Yes, you can do without Fresnel integrals I give you the first step: $$ \int_{y^3}^y \sin\Bigl(\frac{\pi x}{2y}\Bigr)\,dx = \frac{2}{\pi}y\cos\Bigl(\frac{\pi y^2}{2}\Bigr). $$ Then integrate with respect to $y$ (if you do not see how to do that I give you the hint to change variable $u=y^2$).

  • $\begingroup$ oh ) i didnt saw my y before the cos. Fuf, my attention problem. Now its ok, of course ) $\endgroup$ – Ilya.K. Jun 21 '15 at 17:23

See below the range of integration :

enter image description here

  • $\begingroup$ yes, but you did a calculating mistake, when you do integral on cos, it becomes sin, look carefully. My result is $ -sqrt(2)/(pi^2) $ $\endgroup$ – Ilya.K. Jun 21 '15 at 18:20
  • $\begingroup$ How can you find a negative result for the integral of an always positive function ? $\endgroup$ – JJacquelin Jun 21 '15 at 19:02
  • $\begingroup$ Yes, you right, i thought the calculations are right, although it seemed to me suspicious, but i have a calculating mistake. First come logic and then calculations and not the opposite )) you right 100%. My appology. $\endgroup$ – Ilya.K. Jun 21 '15 at 19:56
  • $\begingroup$ All's well that ends well ! $\endgroup$ – JJacquelin Jun 21 '15 at 20:22

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