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

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1like 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1sorry I'm typing this while dealing with a stomachache

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I just typo'ed dropped the x :(

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1u = x^2 du = 2x dx \[\frac{du}{2x} = dx\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so 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) \[\frac{1}{2x}\cos(x^2)+C\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1FORGET IT! It looks innocent, yet when I try to do it, it's all screwed up.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1nugh! there has got to be a way to get rid of that second part!!!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1first part is fine... second is leave cos(x^2) deal with that fraction which is quotient rule

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5hint: we can write this: \[\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \frac{1}{2}\int_{  \infty }^{ + \infty } {\sin } \left( {{x^2}} \right)\;dx\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5now I change variable: \[\large {x^2} = z\] where z is the new integration variable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is no antiderivative 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.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5\[\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \int_0^{ + \infty } {\sin \left( z \right)} \;\frac{{dz}}{{2\sqrt z }}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In case you don't wanna use complex analysis. Michele is on the right track till now.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5the subsequent function: \[\large \frac{{\sin \left( z \right)}}{{2\sqrt z }}\] is holomorphic in all zplane except z=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you wanna use complex analysis?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5may I apply complex analysis?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5ok! Then the line of integration is: dw:1434534738701:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5using the theorem of residual we can write: \[\large \oint {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5next 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5the contribute along the curve in C_0 is zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0On the circles integral is zero right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5yes! that's right!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5next we can write: \[\begin{gathered} \int\limits_{I + } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz + } \int\limits_{I  } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} = \hfill \\ \hfill \\ = \int_0^{ + \infty } {\sin \left( {{x^2}} \right)dx} + \int_{  \infty }^0 {\sin \left( {{x^2}} \right)dx} = \hfill \\ \hfill \\ = 2\int_0^{ + \infty } {\sin \left( {{x^2}} \right)dx} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5so the requested integral is zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think on one of the circles integral has to have some value, because the answer is\[\sqrt{\frac{\pi}{8}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looking 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 \(e^{ix^{2}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Concentrationalizing that will lead us to answer also, quite right

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5the function: \[{\frac{{\sin \left( z \right)}}{{2\sqrt z }}}\] has not poles in zplane, it has only a singularity at z=0, nevertheless the residual value at z=0, is 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \(\frac{{\sqrt{\pi}}}{2\sqrt{2}}\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5we have to evaluate this integral: \[\oint_{{C_0}} {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} \ne 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5Please wait, I have to go out, now, I will return on that question!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5evidently 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, discussion is really good

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm waiting for a solution without complex analysis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll post my solution later.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \(\mathcal{L}\left\{x^{1/2}\right\}=\sqrt{\dfrac{\pi}{s}}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, we used the same method as @Concentrationalizing though we through around the idea of using a contour.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let me know if it is not clear.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0@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.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0very neat solution @mukushla

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tnx @xapproachesinfinity
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