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Calculus Level 5

∫ 0 ∞ x 3 e − 3 x sin ⁡ x d x \large \int _{ 0 }^{ \infty } { x }^{ 3 }{ e }^{ -3x }\sin { x } \, dx

If the integral above is the form of a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .

This is a part of Who's up to the challenge .


The answer is 661.

Hamza A by Hamza A , 19, USA

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3 solutions

Aritra Das
Mar 9, 2016

We just look at the imaginary part of the integral: I = ∫ 0 ∞ x 3 e − 3 x e i x d x = ∫ 0 ∞ x 3 e − ( 3 − i ) x d x I= \int_0^{\infty} x^3 e^{-3x} e^{ix} dx = \int_0^{\infty} x^3 e^{-(3-i)x} dx

With the substitution ( 3 − i ) x = t (3-i)x=t and ( 3 − i ) d x = d t (3-i)dx = dt , we get: I = 1 ( 3 − i ) 4 ∫ 0 ∞ t 3 e − t d t = 7 + 24 i 2500 Γ ( 4 ) I=\frac{1}{(3-i)^4} \int_0^{\infty} t^3 e^{-t} dt = \frac{7+24i}{2500} \Gamma(4)

Using the fact that for integers, Γ ( n ) = ( n − 1 ) ! \Gamma(n) = (n-1)! , the imaginary part of the integral is 24 × 6 2500 = 36 625 \frac{24 \times 6}{2500} = \frac{36}{625} . Hence, a + b = 661 a+b=661

11 Helpful 0 Interesting 1 Brilliant 0 Confused

Same solution!!

Aakash Khandelwal - 5 years, 3 months ago

Hello Aritra , how did you substitute sin x by e^ix , and why did you find only imaginary part of the integral

Ujjwal Mani Tripathi - 5 years, 2 months ago

Log in to reply

e i x = cos ⁡ x + i sin ⁡ x e^{ix} = \cos x + i \sin x , Since integration is linear, the real part of the integrand gives the real part of the result and the same also holds for the imaginary part.

Aritra Das - 5 years, 2 months ago

is doing ( 3 + i ) ∗ ∞ = ∞ (3+i)*\infty = \infty\ ok?

neelesh vij - 3 years, 11 months ago

Let the integral be I I . Then we have:

I = ∫ 0 ∞ x 3 e − 3 x sin ⁡ x d x = ∫ 0 ∞ ∂ 3 ∂ u 3 ( e − 3 x u − 27 ) sin ⁡ x d x where u = 1 = − 1 27 ˙ ∂ 3 ∂ u 3 ∫ 0 ∞ e − 3 x u sin ⁡ x d x Let t = 3 x u = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 3 u ∫ 0 ∞ e − t sin ⁡ ( t 3 u ) d t ) By Maclaurin series: = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 3 u ∫ 0 ∞ e − t ∑ n = 0 ∞ ( − 1 ) n ( t 3 u ) 2 n + 1 ( 2 n + 1 ) ! d t ) = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 3 u ∑ n = 0 ∞ ( − 1 ) n ∫ 0 ∞ t 2 n + 1 e − t d t ( 3 u ) 2 n + 1 ( 2 n + 1 ) ! ) = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 3 u ∑ n = 0 ∞ ( − 1 ) n Γ ( 2 n + 2 ) ( 3 u ) 2 n + 1 ( 2 n + 1 ) ! ) = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 3 u ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ( 3 u ) 2 n + 1 ( 2 n + 1 ) ! ) = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 ( 3 u ) 2 ∑ n = 0 ∞ ( − 1 ( 3 u ) 2 ) n ) = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 ( 3 u ) 2 ( 1 1 + 1 ( 3 u ) 2 ) ) = − 1 27 ˙ ∂ 3 ∂ u 3 ( 1 9 u 2 + 1 ) = 1 27 ( 1944 ( 9 u 2 − 1 ) ( 9 u 2 + 1 ) 4 ) Putting u = 1 = 36 625 \begin{aligned} I & = \int_0^\infty \color{#3D99F6}{x^3 e^{-3x}} \sin x \space dx \\ & = \int_0^\infty \color{#3D99F6}{\frac{\partial^3}{\partial u^3} \left(\frac{e^{-3xu}}{-27}\right)}\sin x \space dx \quad \quad \small \color{#3D99F6}{\text{where }u=1} \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \int_0^\infty e^{-3xu} \sin x \space dx \quad \quad \small \color{#3D99F6}{\text{Let }t = 3xu} \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left( \frac{1}{3u} \int_0^\infty e^{-t} \color{#3D99F6}{\sin \left(\frac{t}{3u}\right)} \space dt \right) \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series:}} \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left( \frac{1}{3u} \int_0^\infty e^{-t} \color{#3D99F6}{\sum_{n=0}^\infty \frac{(-1)^n \left(\frac{t}{3u}\right)^{2n+1}}{(2n+1)!}} \space dt \right) \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left( \frac{1}{3u} \sum_{n=0}^\infty \frac{(-1)^n \int_0^\infty t^{2n+1}e^{-t}\space dt}{(3u)^{2n+1}(2n+1)!} \right) \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left( \frac{1}{3u} \sum_{n=0}^\infty \frac{(-1)^n \Gamma (2n+2)}{(3u)^{2n+1}(2n+1)!} \right) \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left( \frac{1}{3u} \sum_{n=0}^\infty \frac{(-1)^n (2n+1)!}{(3u)^{2n+1}(2n+1)!} \right) \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left( \frac{1}{(3u)^2} \sum_{n=0}^\infty \left(\frac{-1}{(3u)^2} \right)^n \right) \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left( \frac{1}{(3u)^2} \left(\frac{1} {1+\frac{1}{(3u)^2}} \right) \right) \\ & = - \frac{1}{27} \dot{} \frac{\partial^3}{\partial u^3} \left(\frac{1} {9u^2+1} \right) \\ & = \frac{1}{27} \left(\frac{1944(9u^2-1)}{(9u^2+1)^4} \right) \quad \quad \small \color{#3D99F6}{\text{Putting } u = 1} \\ & = \frac{36}{625} \end{aligned}

⇒ a + b = 36 + 625 = 661 \Rightarrow a + b = 36 + 625 = \boxed{661}

12 Helpful 0 Interesting 0 Brilliant 0 Confused
Amol Pai
Mar 13, 2017

We have

I = ∫ 0 ∞ x 3 e − 3 x s i n x d x I=\int_{0}^{\infty} x^3 e^{-3x} sinx dx

Thus, we write

I = ∫ 0 ∞ x 3 e − a x s i n x d x I=\int_{0}^{\infty} x^3 e^{-ax} sinx dx where a=3

Integrating thrice with respect to a, we get

I ′ ′ ′ = − ∫ 0 ∞ e − a x s i n x d x I'''=-\int_{0}^{\infty} e^{-ax} sinx dx

Applying parts by taking e − a x e^{-ax} as the first function, we get

I ′ ′ ′ = − 1 1 + a 2 I'''= \frac{-1}{1+a^2}

Finally, differentiating thrice with respect to a, we get

I = 24 a ( a 2 − 1 ) ( 1 + a 2 ) 4 I= \frac{24a(a^2-1)}{(1+a^2)^4}

Putting a=3, we get

I = 36 625 I=\frac{36}{625}

Thus, a=36, b=625, a+b= 661 \boxed{661}

6 Helpful 0 Interesting 1 Brilliant 0 Confused

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