Paneer Definite Masala 10

Calculus Level 5

I 1 = ∫ 0 π sinc ( x ) d x I 2 = ∫ 0 π / 2 sinc ( x ) sinc ( π 2 − x ) d x \large I_1 =\int_0^\pi \text{sinc}(x) \, dx \qquad I_2 = \int_0^{\pi /2} \text{sinc}(x) \text{sinc} \left( \dfrac\pi2 - x\right) \, dx

I 1 I_1 and I 2 I_2 are two definite integrals as described above and I 1 I 2 = A π B C \dfrac{I_1}{I_2} = \dfrac {A\pi^B}C , where A A , B B and C C are positive integers, find A + B + C A+B+C .

Notation: sinc ( x ) = sin ⁡ x x \text{sinc}(x) = \dfrac{\sin x}x denotes the sinc function .


The answer is 4.

Kishore S. Shenoy by Kishore S. Shenoy , 22, India

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2 solutions

Solution suggested @Aniket Sanghi .

I 1 = ∫ 0 π sin ⁡ x x d x Using ∫ a b f ( a + b − x ) d x = ∫ a b f ( x ) d x = 1 2 ∫ 0 π sin ⁡ x x + sin ⁡ ( π − x ) π − x d x Note that sin ⁡ ( π − x ) = sin ⁡ x = 1 2 ∫ 0 π sin ⁡ x x + sin ⁡ x π − x d x = 1 2 ∫ 0 π π sin ⁡ x x ( π − x ) d x Let x = 2 u ⟹ d x = 2 d u = 1 2 ∫ 0 π 2 2 π sin ⁡ ( 2 u ) 2 u ( π − 2 u ) d u = π 2 ∫ 0 π 2 2 sin ⁡ u cos ⁡ u u ( π − 2 u ) d u = π 2 ∫ 0 π 2 sin ⁡ u sin ⁡ ( π 2 − u ) u ( π 2 − u ) d u = π 2 I 2 ⟹ I 1 I 2 = π 2 \begin{aligned} I_1 & = \int_0^\pi \frac {\sin x}x dx & \small {\color{#3D99F6}\text{Using }\int_a^b f(a+b-x) \ dx = \int_a^b f(x) \ dx} \\ & = \frac 12 \int_0^\pi \frac {\sin x}x + \frac {\sin (\pi - x)}{\pi - x} dx & \small {\color{#3D99F6}\text{Note that } \sin (\pi - x) = \sin x} \\ & = \frac 12 \int_0^\pi \frac {\sin x}x + \frac {\sin x}{\pi - x} dx \\ & = \frac 12 \int_0^\pi \frac {\pi \sin x}{x(\pi - x)} dx & \small {\color{#3D99F6}\text{Let } x= 2u \implies dx = 2 \ du} \\ & = \frac 12 \int_0^\frac \pi 2 \frac {2\pi \sin (2u)}{2u(\pi - 2u)} du \\ & = \frac \pi 2 \int_0^\frac \pi 2 \frac {2\sin u \cos u}{u(\pi - 2u)} du \\ & = \frac \pi 2 \int_0^\frac \pi 2 \frac {\sin u \sin \left(\frac \pi 2 - u \right)}{u \left(\frac \pi 2 - u \right)} du \\ & = \frac \pi 2 I_2 \\ \implies \frac {I_1}{I_2} & = \frac \pi 2 \end{aligned}

⟹ A + B + C = 1 + 1 + 2 = 4 \implies A + B + C = 1+1+2 = \boxed{4}

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Yeah! Thanks sir , for changing my process to complete solution :) !! I actually was busy at that time .

Aniket Sanghi - 4 years, 7 months ago
Aniket Sanghi
Nov 9, 2016

Process :

  • First apply f ( a + b - x) = f (x) to I 1 {I}_{1} (where a and b are limits) and add the 2 terms.
  • Then substitute x = 2t.
  • And You will get the answer.

I 1 I 2 = π 2 \boxed {\frac {{I}_{1}}{{I}_{2}} = \frac { \pi }{2} }

1 Helpful 1 Interesting 1 Brilliant 0 Confused

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