$$\int_0^{+\infty}\sin x^2dx$$
My work:
Let $x^2=t$. Then $\int_0^{+\infty}\sin x^2dx=\int_0^{+\infty}t\sin tdt$
Is it correct? How I will can finish it?
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2$dt =2xdx=2\sqrt t dx$ and $dx=\frac 1 {2\sqrt t} dt$. – Kavi Rama Murthy Nov 22 '19 at 09:44
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3Do you want to prove that converges only or calculate it also? – Marios Gretsas Nov 22 '19 at 09:44
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3$$x^2=t\implies 2x,dx=dt\implies dx=\frac 1{2\sqrt t}dt$$ Your substitution is incorrect – DonAntonio Nov 22 '19 at 09:45
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For explicit calculation look at Fresnel integrals..You will need contour integration and complex analysis. – Marios Gretsas Nov 22 '19 at 09:48
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Is this integral convergent at all ? The function is extremely oscillating for large $x$. – Peter Nov 22 '19 at 09:49
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@Peter The function oscillates more quickly with large $x$, but the oscillations of $\int_0^M \sin (x^2) ,dx$ become smaller as $M$ increases. – Travis Willse Nov 22 '19 at 09:51
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@Peter $sin(x^2)$ oscillates a lot, but that doesn't mean that the integral will. As x increases, the oscillations get narrower, but same amplitude, so the fluctuations in area get smaller. – Sanjay Manohar Nov 22 '19 at 09:52
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Why not you use complex analysis to calculate the integral of $sin(x^2)$ from $0$ to $\infty$. – Emon Hossain Nov 22 '19 at 10:15
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See here for a generalized version: https://math.stackexchange.com/a/2991233/463895 which states the explicit value of the integral – Rohan Shinde Nov 22 '19 at 11:08
1 Answers
2
This is a Fresnel integral. It equals to $\sqrt{\frac{\pi}{8}}$.
$\int_{0}^{+\infty}\sin(x^2)dx = \int_{0}^{1}\sin(x^2)dx + \int_{1}^{+\infty}\sin(x^2)dx$.
The first integral is obviously bounded.
The second integral can be rewritten (as pointed out above in the comments) as
$\int_{1}^{+\infty}\sin(x^2)dx = \left|\begin{array}{c} x^2 = t \Rightarrow 2xdx = dt \\ dx = \frac{dt}{2x} = \frac{dt}{2\sqrt{t}}\end{array}\right| = \int_{1}^{+\infty}\sin(t)\frac{1}{2\sqrt{t}}dt$.
The latter integral converges according to Dirichlet's test:
Let us consider the following integral $\int_a^{+\infty}f(t)g(t)dt$.
In our case,
- $a = 1$,
- $f(t) = \sin(t)$,
- $g(t) = \frac{1}{2\sqrt{t}}$.
1) If the integral of a function $f(t)$ is uniformly bounded over all intervals within $[a, +\infty)$. (TRUE)
2) If $g(t)$ is a monotonically decreasing non-negative function. (TRUE)
Then the integral $\int_a^{+\infty}f(t)g(t)dt$ is a convergent improper integral.
Eugene
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