A calculus problem by Sabhrant Sachan

Calculus Level 5

$\large \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin{x})^{200} \, dx$

The closed form of the integral above can be expressed as $\dfrac{\pi b!}{a^b (c!)^2}$ , where $a,b$ and $c$ are natural numbers. Find $a+b+c$ .

Inspiration

The answer is 302.

2 solutions

Chew-Seong Cheong
Feb 4, 2017

$\begin{aligned} I & = \int_{-\frac \pi 2}^\frac \pi 2 \sin^{200} x \ dx & \small \color{#3D99F6} \text{Since the integral is even} \\ & = 2 \int_0^\frac \pi 2 \sin^{200} x \ dx \\ & = 2 \int_0^\frac \pi 2 \sin^{200} x \cos^0 x \ dx \\ & = B \left(\frac {201}2, \frac 12 \right) & \small \color{#3D99F6} B(m,n) \text{is beta function.} \\ & = \frac {\Gamma \left(\frac {201}2\right)\Gamma \left(\frac 12\right)}{\Gamma \left(101\right)} & \small \color{#3D99F6} \Gamma (s) \text{is gamma function.} \\ & = \frac {\frac {199!!}{2^{100}}\sqrt \pi \sqrt \pi}{100!} \\ & = \frac {200! \pi}{2^{200} (100!)^2} \end{aligned}$

$\implies a + b + c = 2 + 200 + 100 = \boxed{302}$

References:

Beta function
Gamma function

I think it should be

$\begin{aligned} ... \ &= \dfrac{\Gamma\left(\dfrac{201}{2}\right) \Gamma\left(\dfrac{1}{2}\right)}{\Gamma(101)} \\ \\ &= \dfrac{\frac{199\times197\times\cdots\times3}{2^{100}} \sqrt{\pi} \sqrt{\pi}}{100!} \\ \\ \\ &= \dfrac{\frac{200\times199\times198\times197\times\cdots\times3\times2}{2^{100}} \times \frac{1}{2^{100}\times100!}\times\pi}{100!} \\ \\ \\ &= \dfrac{(200)!\pi}{2^{200}{(100!)}^{\color{#D61F06}{2}}} \end{aligned}$

I've added the steps according to my solution. Please make the correction in the final step.

P.S. I don't know what is a double factorial ( $n!!$ ). Can you tell me?

Tapas Mazumdar - 4 years, 4 months ago

$199{\color{#D61F06}!!} = 199 \times 197 \times 195 \times ... \times 1$ . Note that it is double exclamation mark $\color{#D61F06}!!$ .

Chew-Seong Cheong - 4 years, 4 months ago

Oh. So, that represents the factorial function of an AP with common difference 2? Or does it represent the factorial function of an AP consisting of only odd positive integers?

Also, see my comment again, in the last step it should be $\dfrac{(200)!\pi}{2^{200}{(100!)}^{\color{#D61F06}{2}}}$ sir.

Tapas Mazumdar - 4 years, 4 months ago

@Tapas Mazumdar – $8!! = 8 \times 6 \times 4 \times 2$ . Yes, I missed out the square.

Chew-Seong Cheong - 4 years, 4 months ago

@Chew-Seong Cheong – Okay. Thank you sir!

Tapas Mazumdar - 4 years, 4 months ago

I did the same

Sudhamsh Suraj - 4 years, 3 months ago

Prakhar Bindal
Feb 2, 2017

As the given function is even it becomes twice integral from 0 to pi/2 of (sinx)^200

Now use gamma function directly or Walli's Formula

both this and integral inspiration for this were more like the process to get at the asked form was tougher than the integral in general

quite tricky, i felt , but nice :)

Rohith M.Athreya - 4 years, 4 months ago

Yup correct!!. it was a hard time for getting integral Into required form..

Prakhar Bindal - 4 years, 4 months ago

here is another simple one

Rohith M.Athreya - 4 years, 4 months ago

Yes once one gets it to the required form of $B(201/2,1/2)$ we can use Legendre's Duplication formula $z!(z-1/2)!=2^(-2z)√π(2z)!$ .and relationship between Gamma and Beta function.

Spandan Senapati - 4 years, 1 month ago

A calculus problem by Sabhrant Sachan

The answer is 302.

2 solutions

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