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Do I have to integrate the expression by parts and then evaluate the limit to find if the following (improper) integral $$ \int_0^{\infty} \sin(x) \sin(x^2) dx $$ converges?

Please help me out.

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  • $\begingroup$ No. Check for tests for convergence of integrals. $\endgroup$ – SchrodingersCat Jan 16 '17 at 15:01
  • $\begingroup$ @SchrodingersCat Actually, the answer to the question is (given as a hint) "It is $convergent$".... $\endgroup$ – user399078 Jan 16 '17 at 15:02
  • $\begingroup$ have spend at least one of your own thoughts on this question? $\endgroup$ – tired Jan 16 '17 at 15:08
  • $\begingroup$ @Nirbhay I think the "No" has refered to integrating by parts. $\endgroup$ – Wojowu Jan 16 '17 at 15:08
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    $\begingroup$ Your integrand can be written as $1/2(\cos(x^2-x) - \cos(x^2+x)$. It might be easier to show that each of the terms converges individually. E.g. for the second term, you could try the substitution $y=x^2+x$. $\endgroup$ – Fabian Jan 16 '17 at 15:17
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I have filled in all the details, and here they are.

It turns out that the integral does converge.

My method also shows that the integrals $\int_0^{\infty} \sin(x^2\pm x)dx$ converge.

Use $\sin a \sin b = (\cos(a-b)-\cos(a+b))/2$ to show that this depends on $\int_0^{\infty} \cos(x^2\pm x)dx$.

I will now show that these converge by looking at the intervals where $x^2 = 2n \pi$. In these intervals, $x$ is essentially constant, so it is like the integral of cos, which is zero.

Let $x_n^2 = 2\pi n$, so $x_n = \sqrt{2\pi n}$.

Let $I_n =\int_{x_n}^{x_{n+1}} \cos(x^2+cx) dx $, where $|c| \le 1$.

I will show that $|I_n| \lt \dfrac{1}{2\pi n^2} $, so $\sum_{n=1}^{\infty} I_n =\int_{0}^{\infty} \cos(x^2+cx)dx $ converges.

Let $y = x^2$, so $dy = 2x\ dx$ or $dx = dy/(2x) =dy/(2\sqrt{y} $. Let $y_n =\sqrt{x_n} =2\pi n $.

Then $I_n =\int_{y_n}^{y_{n+1}} \dfrac{\cos(y+c\sqrt{y})}{2\sqrt{y}}dy $.

Let's see how the parts of $I_n$ vary.

$y_{n+1} =y_n+2\pi $ so

$\begin{array}\\ \sqrt{y_{n+1}} &=\sqrt{y_n+2\pi}\\ &=\sqrt{y_n}\sqrt{1+2\pi/y_n}\\ &<\sqrt{y_n}(1+\pi/y_n)\\ &=\sqrt{y_n}+\dfrac{\pi}{\sqrt{y_n}}\\ \text{and}\\ \dfrac1{\sqrt{y_{n+1}}} &=\dfrac1{\sqrt{y_n}}(1+2\pi/y_n)^{-1/2}\\ &<\dfrac1{\sqrt{y_n}}(1-\dfrac{\pi}{y_n})\\ \text{and}\\ \dfrac1{\sqrt{y_{n+1}}} &\gt\dfrac1{\sqrt{y_n}}(1-2\pi/y_n)\\ &=\dfrac1{\sqrt{y_n}}-\dfrac{2\pi}{y_n^{3/2}}\\ \end{array} $

Therefore $\dfrac1{2\sqrt{y_n}}\int_{y_n}^{y_{n+1}} \cos(y+c\sqrt{y})dy \gt I_n \gt \dfrac1{2\sqrt{y_{n+1}}}\int_{y_n}^{y_{n+1}} \cos(y+c\sqrt{y})dy $.

Let $J_n =\int_{y_n}^{y_{n+1}} \cos(y+c\sqrt{y})dy $.

Since $|\cos'(y)| \le 1$, $|\cos(y+c\sqrt{y})-\cos(y)| \le |c\sqrt{y}| $. Since $\int_{y_n}^{y_{n+1}} \cos(y)dy = 0 $ and $|c| \le 1$,

$\begin{array}\\ |J_n| &\le\int_{y_n}^{y_{n+1}} |c\sqrt{y}|dy\\ &=|c|\int_{y_n}^{y_{n+1}} \sqrt{y}dy\\ &=|c|\dfrac{y^{-1/2}}{-1/2}\big|_{y_n}^{y_{n+1}}\\ &=2|c|(y_n^{-1/2}-y_{n+1}^{-1/2})\\ &\lt 2(y_n^{-1/2}-\dfrac1{\sqrt{y_n}}+\dfrac{2\pi}{y_n^{3/2}})\\ &= \dfrac{4\pi}{y_n^{3/2}}\\ \end{array} $

Therefore (finally!)

$\begin{array}\\ |I_n| &\lt \dfrac1{2\sqrt{y_n}} J_n\\ &\lt \dfrac1{2\sqrt{y_n}} \dfrac{4\pi}{y_n^{3/2}}\\ &= \dfrac{2\pi}{y_n^2}\\ &= \dfrac{2\pi}{(2\pi n)^2}\\ &= \dfrac{1}{2\pi n^2}\\ \end{array} $

so the sum of the $I_n$ converges.

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