I have to check the convergence of the improper integral $$ \int_{0}^{\infty} \cos^2x\,dx .$$

I have tried to solve it in the following manner: $$\begin{align}\int_{0}^{\infty}\cos^2x\,dx &= \lim_{B \to \infty} \int_{0}^{B}\cos^2x\,dx \\&= \lim_{B \to \infty} \int_{0}^{B}(1-\sin^2(x))\,dx \\ &= \lim_{B \to \infty}[x]_{0}^{B} -\lim_{B \to \infty} \int_{0}^{B}\sin^2(x)\,dx .\end{align}$$

Now the first term tends to infinity as $B \to \infty,$ so the integral is not convergent.

But unfortunately the answer is given to be convergent. I don't understand how ?

Also I have proven in the same way that $$ \int_{0}^{\infty} (\sin x^2)^2\,dx $$ is divergent. So basically I think that if my process is wrong then I am wrong in both cases.

Am I wrong ? If so, looking for guidance then. Thank you.

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    $\begingroup$ You cannot do this this way: the first term goes to $+\infty$, but if the second goes to $-\infty$ you have an indeterminate form. $\endgroup$ – Clement C. Oct 1 '18 at 13:43
  • $\begingroup$ Oh ! I see. Then what way should I prefer ? $\endgroup$ – hiren_garai Oct 1 '18 at 13:44
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    $\begingroup$ What you could do, however, is observe that $\cos^2 x$ is periodic with period $\pi$, and $\int_T^{T+\pi} \cos^2 x dx = \frac{\pi}{2}>0$. So... $\endgroup$ – Clement C. Oct 1 '18 at 13:45
  • $\begingroup$ No, it diverges. See my answer. $\endgroup$ – Clement C. Oct 1 '18 at 13:57
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    $\begingroup$ @ hiren_garai Perhaps there is a typo in your text and the $^2$ is meant to stand under the $\cos$. Then we have $\int_0^{\infty } \cos \left(x^2\right) \, dx = \frac{1}{2} \sqrt{\frac{\pi }{2}}$ $\endgroup$ – Dr. Wolfgang Hintze Oct 1 '18 at 14:02

You cannot proceed this way, as you end up with something of the form $$ \int_0^B dx\, \cos^2 x = u_B - v_B $$ and $\lim_{B\to\infty} u_B = \infty$. But if $\lim_{B\to\infty} v_B = \infty$ as well, you have an indeterminate form and cannot conclude.

However, note that $\cos^2$ is perdiodic with period $\pi$, so that for any integer $k$ $$ \int_0^{k\pi} dx\, \cos^2 x = k\int_0^\pi dx\, \cos^2 x = k\cdot \frac{\pi}{2} $$ and therefore $$ \int_0^{B} dx\, \cos^2 x = \int_0^{\lfloor B/\pi\rfloor \pi} dx\, \cos^2 x + \underbrace{\int_{\lfloor B/\pi\rfloor \pi}^B dx\, \cos^2 x}_{\geq 0} \geq \lfloor B/\pi\rfloor\cdot \frac{\pi}{2} \xrightarrow[B\to\infty]{} \infty $$

  • $\begingroup$ Thank you for your answer. In a similar way I can prove the divergence of $$ \int_{0}^{\infty} \text{(sin x²)² dx} $$, right ? $\endgroup$ – hiren_garai Oct 1 '18 at 14:01
  • $\begingroup$ @hiren_garai yes, indeed. $\endgroup$ – Clement C. Oct 1 '18 at 14:02
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    $\begingroup$ I have one more query, can I visualise the problem in the way that it is the integration of a positive term from $0$ to $\infty$ and so the area of the curve will be unbounded and hence the integral is divergent ? It's just to find a rough idea about the problem . @ClementC. $\endgroup$ – hiren_garai Oct 1 '18 at 14:09
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    $\begingroup$ Of course, one may write $$\int_0^L \cos^2(x)\,dx=\frac12 L+\frac14 \sin(2L) $$and conclude. $\endgroup$ – Mark Viola Oct 1 '18 at 14:37
  • $\begingroup$ I got it.@MarkViola. thanks for your time. $\endgroup$ – hiren_garai Oct 1 '18 at 14:48

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