0

I am confused as to how I should solve the integral $\int_{0}^{\infty}\cos(t^2)dt$. The answers suggest using $\int_{0}^{\infty}\cos(2t^2)dt=\sqrt{\frac{\pi}{4}}$. However, I do not understand how was I supposed to realize that. I'm assuming it's somehow related to the integral $\int_{-\infty}^{\infty}e^{-t^2}dt=\sqrt{\pi}$. Any ideas?

amWhy
  • 210,739
kauselis3000
  • 263

1 Answers1

1

First, realize that cos($x$) = cos($-x$) for all $x$, so $\cos(x^2) = \cos(-x^2)$

Next, $\cos(-x^2) = \Re(e^{-ix^2})$, so the integral goes from $$\int_{0}^{\infty}\cos(x^2)\,dx = \Re\left(\int_{0}^{\infty}e^{-ix^2}\,dx\right)$$

Letting $u = \sqrt{ix} \implies dx = \frac{du}{\sqrt{i}}$, our integral becomes $$\Re\left(\int_{0}^{\infty}e^{-ix^2}\,dx\right) = \Re\left(\int_{0}^{\infty}e^{-u^2}\frac{1}{\sqrt{i}} \,du\right) = \Re(\frac{\sqrt{\pi}}{2\sqrt{i}})$$ using the fact that $\int_{0}^{\infty} e^{-x^2} \,dx = \frac{\sqrt{\pi}}{2}$.

Using the fact that $\frac{1}{\sqrt{i}} = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})$ $$\Re(\frac{\sqrt{\pi}}{2\sqrt{i}}) = \frac{\sqrt{\pi}}{2}(\cos(\frac{-\pi}{4})) = \frac{\sqrt{2\pi}}{4}$$

Thomas Wu
  • 42
  • 5
    The substitution $u := \sqrt{ix}$ introduces a substantial change in the limits of integration, something to address. – A rural reader Jan 21 '23 at 02:37
  • 2
    Please do not answer questions that are duplicates. If your answer is substantially different from those given for the original question, please submit your answer there instead. – Gary Jan 21 '23 at 02:49
  • 1
    The limits after the substitution is $$\lim_{R\to\infty}\int_{-iR}^{iR}f(u),du,$$ not $$\lim\int_{-R}^R f(u),du.$$ Very different thing. – Thomas Andrews Jan 21 '23 at 03:34
  • 2
    On the substitution step, I don't see how $e^{-u^2} = e^{-ix^2}$ when $u=\sqrt{ix}$ – Anas Khaled Jan 21 '23 at 09:06
  • 2
    Yeah, I think Wu wanted to substitute just $u=x\sqrt{i} ,$ given the formula for $du.$ @AnasKhaled But the answer would still be wrong. – Thomas Andrews Jan 22 '23 at 08:03