anonymous
  • anonymous
Integral challenge\[\int_{0}^{\infty} \sin (x^2) \ \text{d}x = ?\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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rational
  • rational
.
UsukiDoll
  • UsukiDoll
u substitution?
UsukiDoll
  • 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

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UsukiDoll
  • UsukiDoll
sorry I'm typing this while dealing with a stomachache
UsukiDoll
  • UsukiDoll
I just typo'ed dropped the x :(
UsukiDoll
  • UsukiDoll
u = x^2 du = 2x dx \[\frac{du}{2x} = dx\]
UsukiDoll
  • 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) \[\frac{-1}{2x}\cos(x^2)+C\]
UsukiDoll
  • UsukiDoll
FORGET IT! It looks innocent, yet when I try to do it, it's all screwed up.
anonymous
  • anonymous
1 Attachment
UsukiDoll
  • UsukiDoll
nugh! there has got to be a way to get rid of that second part!!!
UsukiDoll
  • UsukiDoll
first part is fine... second is leave cos(x^2) deal with that fraction which is quotient rule
Michele_Laino
  • Michele_Laino
hint: we can write this: \[\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\sin } \left( {{x^2}} \right)\;dx\]
Michele_Laino
  • Michele_Laino
now I change variable: \[\large {x^2} = z\] where z is the new integration variable
anonymous
  • 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.
Michele_Laino
  • Michele_Laino
\[\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \int_0^{ + \infty } {\sin \left( z \right)} \;\frac{{dz}}{{2\sqrt z }}\]
anonymous
  • anonymous
In case you don't wanna use complex analysis. Michele is on the right track till now.
Michele_Laino
  • Michele_Laino
the subsequent function: \[\large \frac{{\sin \left( z \right)}}{{2\sqrt z }}\] is holomorphic in all z-plane except z=0
anonymous
  • anonymous
you wanna use complex analysis?
Michele_Laino
  • Michele_Laino
may I apply complex analysis?
anonymous
  • anonymous
oh yeah, no problem
Michele_Laino
  • Michele_Laino
ok! Then the line of integration is: |dw:1434534738701:dw|
Michele_Laino
  • Michele_Laino
using the theorem of residual we can write: \[\large \oint {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} = 0\]
Michele_Laino
  • 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|
Michele_Laino
  • Michele_Laino
the contribute along the curve in C_0 is zero
anonymous
  • anonymous
On the circles integral is zero right?
Michele_Laino
  • Michele_Laino
yes! that's right!
Michele_Laino
  • Michele_Laino
next we can write: \[\begin{gathered} \int\limits_{I + } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz + } \int\limits_{I - } {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} = \hfill \\ \hfill \\ = \int_0^{ + \infty } {\sin \left( {{x^2}} \right)dx} + \int_{ - \infty }^0 {\sin \left( {{x^2}} \right)dx} = \hfill \\ \hfill \\ = 2\int_0^{ + \infty } {\sin \left( {{x^2}} \right)dx} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so the requested integral is zero
anonymous
  • anonymous
I think on one of the circles integral has to have some value, because the answer is\[\sqrt{\frac{\pi}{8}}\]
anonymous
  • 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 \(e^{ix^{2}}\)
anonymous
  • anonymous
@Concentrationalizing that will lead us to answer also, quite right
Michele_Laino
  • 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
anonymous
  • 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 \(\frac{{\sqrt{\pi}}}{2\sqrt{2}}\)
Michele_Laino
  • Michele_Laino
we have to evaluate this integral: \[\oint_{{C_0}} {\frac{{\sin \left( z \right)}}{{2\sqrt z }}dz} \ne 0\]
Michele_Laino
  • Michele_Laino
Please wait, I have to go out, now, I will return on that question!
Michele_Laino
  • 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|
perl
  • perl
Nice post.
anonymous
  • anonymous
Yes, discussion is really good
anonymous
  • anonymous
I'm waiting for a solution without complex analysis
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Fresnel_integral
anonymous
  • anonymous
I'll post my solution later.
anonymous
  • 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 \(\mathcal{L}\left\{x^{-1/2}\right\}=\sqrt{\dfrac{\pi}{s}}\).
anonymous
  • anonymous
Actually, we used the same method as @Concentrationalizing though we through around the idea of using a contour.
anonymous
  • anonymous
anonymous
  • anonymous
Let me know if it is not clear.
zepdrix
  • 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.
xapproachesinfinity
  • xapproachesinfinity
very neat solution @mukushla
anonymous
  • anonymous
tnx @xapproachesinfinity

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