## anonymous one year ago Integral challenge$\int_{0}^{\infty} \sin (x^2) \ \text{d}x = ?$

1. rational

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2. UsukiDoll

u substitution?

3. UsukiDoll

like let u = x^2 du = 2dx $\frac{du}{2} = dx$ and then -cos x because the derivative of cos x is -sin x and if we have that extra negative we can multiply . A negative x negative is positive -(-sinx) = sinx

4. UsukiDoll

sorry I'm typing this while dealing with a stomachache

5. UsukiDoll

I just typo'ed dropped the x :(

6. UsukiDoll

u = x^2 du = 2x dx $\frac{du}{2x} = dx$

7. UsukiDoll

so to gain sin(u) back I need -cos(u) because the derivative of cos(u) is -sin(u) and with that extra negative in place... a negative x a negative is a positive so -(-sin(u)) ->sin (u) $\frac{-1}{2x}\cos(x^2)+C$

8. UsukiDoll

FORGET IT! It looks innocent, yet when I try to do it, it's all screwed up.

9. anonymous

10. UsukiDoll

nugh! there has got to be a way to get rid of that second part!!!

11. UsukiDoll

first part is fine... second is leave cos(x^2) deal with that fraction which is quotient rule

12. Michele_Laino

hint: we can write this: $\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\sin } \left( {{x^2}} \right)\;dx$

13. Michele_Laino

now I change variable: $\large {x^2} = z$ where z is the new integration variable

14. anonymous

There is no anti-derivative for $$\sin x^2$$ in terms of elementary functions. But there is a way to solve this integral with elementary functions. Also a little bit of Laplace transforms needed here and first thing to do is changing it to a double integral.

15. Michele_Laino

$\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \int_0^{ + \infty } {\sin \left( z \right)} \;\frac{{dz}}{{2\sqrt z }}$

16. anonymous

In case you don't wanna use complex analysis. Michele is on the right track till now.

17. Michele_Laino

the subsequent function: $\large \frac{{\sin \left( z \right)}}{{2\sqrt z }}$ is holomorphic in all z-plane except z=0

18. anonymous

you wanna use complex analysis?

19. Michele_Laino

may I apply complex analysis?

20. anonymous

oh yeah, no problem

21. Michele_Laino

ok! Then the line of integration is: |dw:1434534738701:dw|

22. Michele_Laino

using the theorem of residual we can write: $\large \oint {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} = 0$

23. Michele_Laino

next we have: $\Large \oint {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} = \int\limits_{I + } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz + } \int\limits_{I - } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz}$|dw:1434535093608:dw|

24. Michele_Laino

the contribute along the curve in C_0 is zero

25. anonymous

On the circles integral is zero right?

26. Michele_Laino

yes! that's right!

27. Michele_Laino

next we can write: $\begin{gathered} \int\limits_{I + } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz + } \int\limits_{I - } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} = \hfill \\ \hfill \\ = \int_0^{ + \infty } {\sin \left( {{x^2}} \right)dx} + \int_{ - \infty }^0 {\sin \left( {{x^2}} \right)dx} = \hfill \\ \hfill \\ = 2\int_0^{ + \infty } {\sin \left( {{x^2}} \right)dx} \hfill \\ \end{gathered}$

28. Michele_Laino

so the requested integral is zero

29. anonymous

I think on one of the circles integral has to have some value, because the answer is$\sqrt{\frac{\pi}{8}}$

30. anonymous

Looking at it, I guess the way that made sense to me was considering $$cos(x^{2}) + isin(x^{2}) = e^{ix^{2}}$$ and then dealing with the integral of $$e^{ix^{2}}$$

31. anonymous

32. Michele_Laino

the function: ${\frac{{\sin \left( z \right)}}{{2\sqrt z }}}$ has not poles in z-plane, it has only a singularity at z=0, nevertheless the residual value at z=0, is 0

33. anonymous

I wish I perfectly understand this process, but I do know a little bit to type up a jumbled mess. So first note this: $$cos(x^{2}) + isin(x^{2}) = e^{ix^{2}}$$ So instead of looking directly at the integral of $$sin(x^{2})$$, we would want to try and mess with $$e^{ix^{2}}$$. Okay, so $$e^{iz^{2}}$$ is entire, which means $$\int_C e^{iz^{2}} = 0$$ for any closed contour C. For dealing with this integral, we consider this contour |dw:1434535791428:dw| The reason the angle of the chosen contour stops at $$\pi/4$$ is because on the arc of the circle, $$z=re^{i \theta}$$. Choosing $$\theta = \pi/4$$ gives us this nice result: $$e^{iz^{2}} = e^{i(re^{i\pi/4})^{2}} = e^{-r^{2}}$$ And this is nice because $\int\limits_{0}^{+\infty}e^{-x^{2}}dx = \frac{ \sqrt{\pi} }{ 2 }$ So we can add up the integrals along the 3 curves and get a result $\int\limits_{0}^{+\infty}e^{iz^{2}}dz = \lim_{R \rightarrow +\infty} (\int\limits_{0}^{R}e^{ix^{2}}dx + 0 -e^{i\pi/4}\int\limits_{0}^{R}e^{-x^{2}}dx)$ The middle part is 0 because the integral on C2 would just be 0 along that arc. As R goes to $$+\infty$$, The first integral tends to the integral we're looking for while the 3rd integral is a known result. $\int\limits_{0}^{+\infty}e^{iz^{2}}dz = \int\limits_{0}^{+\infty}e^{ix^{2}}dx -(\frac{ 1 }{ \sqrt{2} }+i \frac{ 1 }{ \sqrt{2} })\cdot \frac{ \sqrt{\pi} }{ 2 }$ Of course the left hand side is 0 as previously mentioned. Now we can just move the result of the 3rd integral to the other side and get a result. $\int\limits_{0}^{+\infty}e^{ix^{2}}dx = \frac{ \sqrt{\pi} }{ 2\sqrt{2} } + i \frac{ \sqrt{\pi} }{ 2\sqrt{2} }$ $\int\limits_{0}^{+\infty} \cos(x^{2})dx + i \int\limits_{0}^{+\infty}\sin(x^{2})dx = \frac{ \sqrt{\pi} }{ 2\sqrt{2} } + i \frac{ \sqrt{\pi} }{ 2\sqrt{2} }$ Real parts equal real parts, imaginary parts equal imaginary parts, and we get those both the required integral as well as the one for $$\int_{0}^{+\infty}cos(x^{2})dx$$ give us $$\frac{{\sqrt{\pi}}}{2\sqrt{2}}$$

34. Michele_Laino

we have to evaluate this integral: $\oint_{{C_0}} {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} \ne 0$

35. Michele_Laino

Please wait, I have to go out, now, I will return on that question!

36. Michele_Laino

evidently the values of these integrals: $\oint\limits_{{C_0}} {\frac{{\sin z}}{{2\sqrt z }}} dz,\quad \oint\limits_\Gamma {\frac{{\sin z}}{{2\sqrt z }}} dz$ are not zero |dw:1434559993342:dw|

37. perl

Nice post.

38. anonymous

Yes, discussion is really good

39. anonymous

I'm waiting for a solution without complex analysis

40. anonymous
41. anonymous

I'll post my solution later.

42. anonymous

I actually posted the same question a while back, and @Michele_Laino and I arrived at the answer (more or less) in much the same way. The Laplace transform is a neat way of approaching this, provided we can accept the fact that $$\mathcal{L}\left\{x^{-1/2}\right\}=\sqrt{\dfrac{\pi}{s}}$$.

43. anonymous

Actually, we used the same method as @Concentrationalizing though we through around the idea of using a contour.

44. anonymous

45. anonymous

Let me know if it is not clear.

46. zepdrix

@Concentrationalizing Ohhh thanks for the tips. For some reason I kept thinking the angle should extend from 0 to pi/2. No wonder my final result was missing a 2 under the root.

47. xapproachesinfinity

very neat solution @mukushla

48. anonymous

tnx @xapproachesinfinity