Gamma, Gamma, everywhere! P:3

Calculus Level 4

∫ 0 π 2 sin ⁡ ( 2 x ) sin ⁡ 2 ( x 2 ) d x = Γ 2 ( a b ) π d − π c \int_0^{\frac{π}{2}} \sqrt{\sin(2x)}\sin^2 \left(\frac{x}{2}\right) dx = \frac{\Gamma^2 {\left(\frac{a}{b}\right)}}{\sqrt{πd}} -\frac{π}{c}

a , b , c , d a, b , c, d , are positive integers . Where a a and b b are co-primes .

Find ( 2 d − b ) ( a + c ) (2d-b)(a+c)

Problem is not original


The answer is 0.

Dwaipayan Shikari by Dwaipayan Shikari , 17, India

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1 solution

Dwaipayan Shikari
Dec 11, 2020

∫ 0 π / 2 sin ⁡ 2 x sin ⁡ 2 ( x 2 ) d x = 1 2 ∫ 0 π / 2 s i n x c o s x ( 1 − c o s x ) d x \int_0^{π/2} \sin2x \sin^2 (\frac{x}{2}) dx = \frac{1}{\sqrt{2}} \int_0^{π/2} \sqrt{{sinx }{cosx}}{(1-{cosx})} dx take sin ⁡ 2 x = t \sin^2 x = t then the integral becomes) 1 2 2 ( ∫ 0 1 t − 1 4 ( 1 − t ) − 1 4 − ∫ 0 1 t − 1 4 ( 1 − t ) 1 4 d t ) \frac{1}{2\sqrt{2}}( \int_0^1 t^{-\frac{1}{4}} (1-t)^{-\frac{1}{4}} - \int_0^1 t^{-\frac{1}{4}} (1-t)^{\frac{1}{4}}dt) = 1 2 2 ( B ( 3 4 , 3 4 ) − B ( 3 4 , 5 4 ) ) = \frac{1}{2\sqrt{2}} (\Beta{(\frac{3}{4} ,\frac{3}{4})} - \Beta{(\frac{3}{4} , \frac{5}{4})} ) = Γ 2 ( 3 4 ) 2 2 Γ ( 3 2 ) − Γ ( 3 4 ) Γ ( 5 4 ) 2 2 Γ ( 2 ) = \frac{\Gamma^2{(\frac{3}{4})}}{2\sqrt{2}{\Gamma{(\frac{3}{2})}}} - \frac{\Gamma{(\frac{3}{4})}\Gamma{(\frac{5}{4})}}{2\sqrt{2}\Gamma{(2)}} = Γ 2 ( 3 4 ) 2 π − Γ ( 3 4 ) Γ ( 1 4 ) 8 2 = \frac{\Gamma^2 {(\frac{3}{4})}}{\sqrt{2π}} - \frac{\Gamma{(\frac{3}{4})}\Gamma{(\frac{1}{4})}}{8\sqrt{2}} = Γ 2 ( 3 4 ) 2 π − π 8 2 sin ⁡ ( π 4 ) = \frac{\Gamma^2 {(\frac{3}{4})}}{\sqrt{2π}} -\frac{π}{8\sqrt{2}\sin{(\frac{π}{4})}} Note , Γ ( n ) Γ ( 1 − n ) = π sin ⁡ ( n π ) \color{#E81990}\textrm{Note} , \Gamma{(n)}\Gamma{(1-n)} = \frac{π}{\sin{(nπ)}} ⟹ Γ 2 ( 3 4 ) 2 π − π 8 \implies{\boxed{\frac{\Gamma^2 {(\frac{3}{4})}}{\sqrt{2π}} -\frac{π}{8}}} Answer is ( 2 d − b ) ( a + c ) = 0 \color{#20A900}\boxed{(2d-b)(a+c)=0}

Problem source

1 Helpful 0 Interesting 2 Brilliant 0 Confused

You can use the trigonometric representation of the Beta function -

1 2 ∫ 0 π 2 sin ⁡ x cos ⁡ x − 1 2 ∫ 0 π 2 sin ⁡ 1 2 x cos ⁡ 3 2 x d x = 1 2 2 ( B ( 3 4 , 3 4 ) − B ( 3 4 , 5 4 ) ) \frac {1}{\sqrt 2}\int_{0}^{\frac {\pi}{2}} \sqrt {\sin x \cos x} - \frac {1}{\sqrt 2}\int_{0}^{\frac {\pi}{2}} \sin^{\frac {1}{2}} x \cos^{\frac {3}{2}} x dx = \frac {1}{2\sqrt 2} \left(\Beta (\frac {3}{4}, \frac {3}{4}) - \Beta (\frac {3}{4}, \frac {5}{4})\right)

Since,

B ( x , y ) = 2 ∫ 0 π 2 sin ⁡ 2 x − 1 x ⋅ cos ⁡ 2 y − 1 x d x \Beta (x,y) = 2\int_{0}^{\frac {\pi}{2}} \sin^{2x-1} x \cdot \cos^{2y-1} x dx

N. Aadhaar Murty - 6 months ago

It's not a bad problem...but why put (2d-b) ?. It makes it easier for people who are just guessing the answer as 0. Its better to just ask for a+b+c+d.

Arghyadeep Chatterjee - 3 months, 1 week ago

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Answers to my questions are simple ,but the questions are tricky. :)

Dwaipayan Shikari - 3 months, 1 week ago

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I get that....but I think you should not ask for 2 d − b 2d-b because in order to correctly solve this problem one need not even have to correctly calculate the value of a a and c c . For example someone who has found the answer to be Γ 2 ( 7 4 ) 2 π − π 1729 \displaystyle \frac{\Gamma^{2}(\frac{7}{4})}{\sqrt{2\pi}} -\frac{\pi}{1729} . Would still get the answer to the problem correct because 2 d − b 2d-b would still be 0 0 but that does not mean he did the original problem correctly. Ofcourse it is up to the problem maker's choice and discretion to ask for whatever answer he wishes to be inputted . But I think it is better to ask for values to the answer such that all the parameters are used. If one finds the answer in terms of parameters a , b , c , d a,b,c,d then I think it is wise to ask for an answer which would involve all 4 4 parameters . Otherwise why should anyone even bother with the π c \frac{\pi}{c} term. I agree that your questions are both tricky and good and frankly you are one of the few people who are currently posting good calculus problem. Otherwise the thriving community that brilliant.org was ....is now reduced to only but a shadow of it's former self. Great problem makers like Satyajit Mohanty , Hamza A , Aditya Kumar , Pratik Shastri , Ariel Gershon and the likes have all become dormant . It is good to have problem maker among us in the calculus section who is actively posting great problems. Feel free to try out my own problems and post solutions.

Arghyadeep Chatterjee - 3 months, 1 week ago

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@Arghyadeep Chatterjee – Thanks brother ! I will try to do better on my question formatting.

Dwaipayan Shikari - 3 months, 1 week ago

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