A calculus problem by Rahil Sehgal

Calculus Level 4

∫ 0 ∞ ∫ x ∞ e − y 2 d y d x \large \int_{0}^{∞} \int_{x}^{∞} e^{-y^{2}} dy \; dx

If the multiple integral above can be expressed in the form of a b \dfrac{a}{b} where a a and b b are positive coprime integers, then enter a + b a+b as your answer.


The answer is 3.

Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Brian Moehring
Apr 30, 2017

By Fubini-Tonelli, ∫ 0 ∞ d x ∫ x ∞ e − y 2 d y = ∫ 0 ∞ d y ∫ 0 y e − y 2 d x = ∫ 0 ∞ y e − y 2 d y = 1 2 ∫ 0 ∞ e − u d u = 1 2 . \int_0^\infty dx \int_x^\infty e^{-y^2}dy = \int_0^\infty dy \int_0^y e^{-y^2}dx = \int_0^\infty ye^{-y^2}dy = \frac 12\int_0^\infty e^{-u}du = \frac 12.

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Way simpler than mine! +1!

Guilherme Niedu - 4 years, 1 month ago

The theorem makes the problem much simpler

Aditya Kumar - 3 years, 11 months ago
Guilherme Niedu
Apr 30, 2017

I = ∫ 0 ∞ ∫ x ∞ e − y 2 d y d x \large \displaystyle I = \int_0^{\infty} \int_x^{\infty} e^{-y^2} dy dx

Make b = y − x \color{#20A900} b = y-x , a = x \color{#20A900} a = x . The limits for both a \color{#20A900} a and b \color{#20A900} b will be 0 \color{#20A900} 0 to ∞ \color{#20A900} \infty , and the determinant of the jacobian will be 1 \color{#20A900} 1 . y \color{#20A900} y will be a + b \color{#20A900} a+b . So:

I = ∫ 0 ∞ ∫ 0 ∞ e − ( a + b ) 2 d b d a \large \displaystyle I = \int_0^{\infty} \int_0^{\infty} e^{-(a+b)^2} db da

Using polar coordinates, a = r cos ⁡ θ \color{#20A900} a = r \cos\theta , b = r sin ⁡ θ \color{#20A900} b = r \sin \theta , ∣ J ∣ = r \color{#20A900} |J| = r :

I = ∫ 0 π 2 ∫ 0 ∞ r e − ( r 2 + 2 r 2 cos ⁡ θ sin ⁡ θ ) d r d θ \large \displaystyle I = \int_0^{\frac{\pi}{2}} \int_0^{\infty} r e^{-(r^2 + 2r^2\cos \theta\sin \theta ) } dr d \theta

I = ∫ 0 π 2 ∫ 0 ∞ r e − r 2 ( 1 + 2 cos ⁡ θ sin ⁡ θ ) d r d θ \large \displaystyle I = \int_0^{\frac{\pi}{2}} \int_0^{\infty} r e^{-r^2 (1 + 2 \cos \theta\sin \theta) } dr d \theta

I = ∫ 0 π 2 [ e − r 2 ( 1 + 2 cos ⁡ θ sin ⁡ θ ) − 2 ( 1 + 2 cos ⁡ θ sin ⁡ θ ) ] 0 ∞ d θ \large \displaystyle I = \int_0^{\frac{\pi}{2}} \left [ \frac{e^{-r^2 (1 + 2 \cos \theta\sin \theta) }}{-2 (1 + 2 \cos \theta\sin \theta) } \right ]_0^{\infty} d\theta

I = 1 2 ∫ 0 π 2 1 1 + 2 cos ⁡ θ sin ⁡ θ d θ \large \displaystyle I = \frac12 \int_0^{\frac{\pi}{2}} \frac{1}{1+ 2 \cos \theta\sin \theta} d\theta .

Make u = tan ⁡ θ \color{#20A900} u = \tan \theta . Then cos ⁡ θ = 1 u 2 + 1 \color{#20A900} \cos \theta = \frac{1}{\sqrt{u^2+1}} , sin ⁡ θ = u u 2 + 1 \color{#20A900} \sin \theta = \frac{u}{\sqrt{u^2+1}} , d u = s e c 2 θ d θ → d θ = d u cos ⁡ 2 θ = d u u 2 + 1 \color{#20A900} du = sec^2 \theta d\theta \rightarrow d \theta = du \cos^2 \theta = \frac{du}{u^2+1} . So:

I = 1 2 ∫ 0 ∞ d u ( u 2 + 1 ) ( 1 + 2 1 u 2 + 1 u u 2 + 1 ) d u \large \displaystyle I = \frac12 \int_0^{\infty} \frac{du}{(u^2+1)(1 + 2 \frac{1}{\sqrt{u^2+1}} \frac{u}{\sqrt{u^2+1}} )} du

I = 1 2 ∫ 0 ∞ d u ( u 2 + 1 ) ( 1 + 2 u u 2 + 1 ) d u \large \displaystyle I = \frac12 \int_0^{\infty} \frac{du}{ (u^2+1)( 1 + 2 \frac{u}{u^2+1} ) } du

I = 1 2 ∫ 0 ∞ d u u 2 + 2 u + 1 d u \large \displaystyle I = \frac12 \int_0^{\infty} \frac{du}{u^2 + 2u + 1}du

I = 1 2 ∫ 0 ∞ d u ( u + 1 ) 2 d u \large \displaystyle I = \frac12 \int_0^{\infty} \frac{du}{ (u+1)^2} du

I = 1 2 [ − 1 u + 1 ] 0 ∞ \large \displaystyle I = \frac12 \left[ -\frac{1}{u+1} \right ] _0^{\infty}

I = 1 2 \color{#20A900} \boxed{ \large \displaystyle I = \frac12 }

Then:

a = 1 , b = 2 , a + b = 3 \color{#3D99F6} \large \displaystyle a=1, b=2, \boxed{ \large \displaystyle a+b=3}

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