Inspired by Jose Sacramento

Calculus Level 5

$\lim_{(a,b)\to \infty}\int_{0}^{a}\int_{0}^{b}\sin\left(\frac{\pi x^2}{2}+\frac{\pi y^2}{2}\right)\, dy\; dx$

Evaluate this limit, rounded to three significant figures.

If you come to the conclusion that this limit fails to exist, enter 666.

Formal Definition : $\displaystyle \lim_{(a,b)\to\infty}f(a,b)=L$ means that for every $\epsilon>0$ there exists an $N$ such that $|f(a,b)-L|<\epsilon$ whenever $a>N$ and $b>N$ .

Inspiration

The answer is 0.5.

1 solution

Ayush Verma
Mar 20, 2016

$as\quad \int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \sin { \left( \cfrac { \pi { x }^{ 2 } }{ 2 } \right) } \cos { \left( \cfrac { \pi y^{ 2 } }{ 2 } \right) } dydx } } =\int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \cos { \left( \cfrac { \pi { x }^{ 2 } }{ 2 } \right) } \sin { \left( \cfrac { \pi y^{ 2 } }{ 2 } \right) } dydx } } \\ \\ I=\int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \sin { \left( \cfrac { \pi { x }^{ 2 } }{ 2 } +\cfrac { \pi { y }^{ 2 } }{ 2 } \right) } dydx } } =2\int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \sin { \left( \cfrac { \pi { x }^{ 2 } }{ 2 } \right) } \cos { \left( \cfrac { \pi y^{ 2 } }{ 2 } \right) } dydx } } \\ \\ =2\left\{ \int _{ 0 }^{ \infty }{ \cos { \left( \cfrac { \pi y^{ 2 } }{ 2 } \right) } dy } \right\} \left\{ \int _{ 0 }^{ \infty }{ \sin { \left( \cfrac { \pi { x }^{ 2 } }{ 2 } \right) } dx } \right\} \\ \\ now,\quad \int _{ 0 }^{ \infty }{ \cos { \left( \cfrac { \pi y^{ 2 } }{ 2 } \right) } dy } =\int _{ 0 }^{ \infty }{ \cfrac { \cos { u } }{ \sqrt { 2\pi u } } dz } \quad \left( if\cfrac { \pi y^{ 2 } }{ 2 } =u\quad \right) \\ \\ =\cfrac { 1 }{ 2 } \\ \\ \\ \& \int _{ 0 }^{ \infty }{ \sin { \left( \cfrac { \pi { x }^{ 2 } }{ 2 } \right) } dx } =\int _{ 0 }^{ \infty }{ \cfrac { \sin { v } }{ \sqrt { 2\pi v } } dz } \quad \left( if\cfrac { \pi x^{ 2 } }{ 2 } =v\quad \right) \\ \\ =\cfrac { 1 }{ 2 } \\ \\ \therefore \quad I=2\left( \cfrac { 1 }{ 2 } \right) \left( \cfrac { 1 }{ 2 } \right) =0.5$

Yes, exactly! (+1) The Dirichlet Kernel is key.

Otto Bretscher - 5 years, 2 months ago

Where is the working for Dirichlet Kernel? All I see is Fresnel integral.

Pi Han Goh - 5 years, 2 months ago

The two are closely related, of course.

Otto Bretscher - 5 years, 2 months ago

@Otto Bretscher – How so? I don't see any word "Fresnel" in wikipedia: Dirichlet kernel .

Pi Han Goh - 5 years, 2 months ago

@Pi Han Goh – There is a way to derive the limits of the Fresnel integral from the Dirichlet kernel (if you don't want to use complex contours); I will find you a reference when I get around to it. (School is back in session now, after Spring Break, and I have less time to "play.")

Otto Bretscher - 5 years, 2 months ago

So then, how is it that the insipiration problem has a diverging integral?

A Former Brilliant Member - 5 years, 2 months ago

wait... I don't understand. but what's wrong with this approach?

$\int_0^\infty \int_0^\infty \sin\left(\frac{\pi}{2}(x^2+y^2)\right)\,dx\,dy=\int_0^\frac{\pi}{4} \int_0^\infty \sin\left(\frac{\pi r^2}{2}\right)r\,dr\,d\theta=\frac{\pi}{4} \int_0^\infty \sin\left(\frac{\pi r^2}{2}\right)r\,dr=\text{does not exist}$

Ruofeng Liu - 1 year, 6 months ago

Inspired by Jose Sacramento

The answer is 0.5.

1 solution

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