When integrals meet abominable numbers

Calculus Level 5

$\int_0^{\frac \pi 2 } \bigg [ ( \sin^{12} x ) \ln (\sin x) \bigg ] \ dx = \frac {\pi}{A \cdot 2^B} \ (C - D \ln 2)$

The equation above is true for integer constants $A,B,C,D$ . Find sum of the sum of digits of $A,B,C$ and $D$

Hints:

$A$ , $B$ are $2$ -digit numbers which differ by $1$ . $C$ and $D$ are 5 digit numbers of which $C$ has only two prime factors.

Comments:

This question may be extremely difficult/calculative if done in certain ways. But a general method may be devised.... NOT for the light hearted.

The answer is 46.

2 solutions

Raghav Vaidyanathan
Feb 19, 2015

The expression is: $\frac { \pi }{ 15{ \times 2 }^{ 14 } } \times \left( 18107-27720\ln { 2 } \right) = -0.01415148...$

consider: $I(n)=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ n }{ x } } dx$ for even $n$

$I(n)=\frac { (n-1)(n-3)(n-5)...(1) }{ (n)(n-2)(n-4)... } \times \frac { \pi }{ 2 }$

By Walli's formula.

It can be shown that:

$\frac { dI(n) }{ dn } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ n }{ x } \log { x } } dx$

$\frac { dI(n) }{ dn }=I(n)\left[ (\frac { 1 }{ n-1 } +\frac { 1 }{ n-3 } +\frac { 1 }{ n-5 } +...(1))-(\frac { 1 }{ n } +\frac { 1 }{ n-2 } +\frac { 1 }{ n-4 } +...+\frac { 1 }{ 2 } )-\log { 2 } \right]$

I got the above expression by differentiating after taking logarithm in the expression for $I(n)$ .

Just put in $n=12$ to get required answer.

Hats off to you dude !!!! Consistently posting Level 5 questions is something not everyone is capable of :)

A Former Brilliant Member - 6 years, 3 months ago

Thank you Roopesh. A lot of thought went into these questions. Good to know that people appreciate it.

Raghav Vaidyanathan - 6 years, 3 months ago

How did you arrive till this part..??

Vighnesh Raut - 6 years, 3 months ago

Bringing the answer into this form was hard,also how did you calculate value of digamma at 13/2

Akshay Bodhare - 6 years, 3 months ago

I spent a lot of time on this question of Ronak Agarwal, which, as of now is worth 400 points. I came across the property I used in my question while solving it. I didn't understand Ronak's solution as it involved a lot of stuff that I haven't learnt. So I put this up, something that people with basic knowledge of integration can also solve. I looked up digamma function. It looks like it's a much more formal version of what I did here. And yes, I used a calculator to get the ugly answer into this form.

Raghav Vaidyanathan - 6 years, 3 months ago

Generally, $\int_0^{\pi/2}\sin^u x \log {(\sin x)} dx = \frac {1}{4} B(\frac {1+u}{2},\frac {1}{2})(\psi (\frac {1+u}{2})-\psi (1+\frac {u}{2}))$

Akshay Bodhare - 6 years, 3 months ago

Incredible Mind
Mar 1, 2015

i too used digamma function...

When integrals meet abominable numbers

The answer is 46.

2 solutions

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