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

- anonymous

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

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- rational

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- UsukiDoll

u substitution?

- 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

sorry I'm typing this while dealing with a stomachache

- UsukiDoll

I just typo'ed dropped the x :(

- UsukiDoll

u = x^2
du = 2x dx
\[\frac{du}{2x} = dx\]

- 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

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

- anonymous

##### 1 Attachment

- UsukiDoll

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

- UsukiDoll

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

- 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

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

- 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

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

- anonymous

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

- Michele_Laino

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

- anonymous

you wanna use complex analysis?

- Michele_Laino

may I apply complex analysis?

- anonymous

oh yeah, no problem

- Michele_Laino

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

- Michele_Laino

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

- 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

the contribute along the curve in C_0 is zero

- anonymous

On the circles integral is zero right?

- Michele_Laino

yes! that's right!

- 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

so the requested integral is zero

- anonymous

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

- 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

@Concentrationalizing that will lead us to answer also, quite right

- 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

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

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

- Michele_Laino

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

- 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

Nice post.

- anonymous

Yes, discussion is really good

- anonymous

I'm waiting for a solution without complex analysis

- anonymous

https://en.wikipedia.org/wiki/Fresnel_integral

- anonymous

I'll post my solution later.

- 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

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

- anonymous

##### 1 Attachment

- anonymous

Let me know if it is not clear.

- 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

very neat solution @mukushla

- anonymous

tnx @xapproachesinfinity

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