Question

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\[\int_0^\infty \cos(x^2)\,dx=\int_0^\infty \sin(x^2)\,dx=\sqrt{\frac{\pi}{8}}\]

Answer 1

[ A relatively simple problem compared to the recent ones :) ]

Answer 2

I can write the result as below:
\[\Im \left( \int\limits_{0}^{+\infty} e ^{ix ^{2}} dx\right)\]
then I use this substitution:
\[ix ^{2}=y\]
and observing that \[\sqrt{i}=\frac{ 1 }{ \sqrt{2} }(1\pm i)\]
I got this:\[-\frac{ 1 }{ 2 \sqrt{i} }\Gamma \left( \frac{ 1 }{ 2 } \right)\]
so the final result is:
\[\pm \frac{ \sqrt{\pi} }{ 2\sqrt{2} }\]

Answer 3

where:\[\Im \]
stands for imaginary part of a complex number

Answer 4

oops...I used:
\[ix ^{2}=-y\]

Answer 5

Well done, I had a similar method, but I keep tripping up with the complex computation:
\[\begin{align*}
\int_{-\infty}^\infty e^{-x^2}\,dx&=\int_{-\infty}^\infty e^{(ix)^2}\,dx\\\\
\sqrt\pi&=\int_{-\infty}^\infty\bigg(\cos (ix)^2+i\sin(ix)^2\bigg)\,dx\\\\
&=\int_{-\infty}^\infty\cos(x^2)\,dx-\underbrace{\mathrm{Im}\left[\int_{-\infty}^\infty e^{-ix^2}\,dx\right]}\end{align*}\]
The bracketed part is the part that's particularly bothersome. At any rate, once I figure this out I'd divide by 2 and the result follows.

Answer 6

I get that the bracketed part amounts to \(-\dfrac{\sqrt{\pi}}{\sqrt2}\), so I suppose there's a mistake in my reasoning somewhere along the line...

Answer 7

It's probably due to
\[\large\int_\mathbb{R}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}\quad\text{for Re}(a)>0\]
and not for complex \(a\).

Answer 8

you are right! @SithsAndGiggles

Answer 9

*not for \(\text{Re}(a)\le0\), I mean

Answer 10

I think so!

Answer 11

very nice question :) @SithsAndGiggles

Answer 12

Thanks! I originally considered approaching the integral via differentiation under the integral sign but I couldn't set up the proper parameterization.

Answer 13

Here's an interesting approach: http://math.stackexchange.com/questions/187729/evaluating-int-0-infty-sin-x2-dx-with-real-methods/190293#190293

Answer 14

I understand

Answer 15

I see

Answer 16

thanks! for your website @SithsAndGiggles