You must have solved the integral of log of cosine

Calculus Level 5

I = 0 π / 2 x 2 log ( sec ( x ) ) d x I=\displaystyle \int _{ 0 }^{ \pi /2 }{ { x }^{ 2 }\log ( \sec { (x) } ) dx }

If the value of I I can be represented as = π A ζ ( B ) + π C log D E \dfrac{\pi}{A} \zeta{(B)} + \dfrac{{\pi}^{C} \log{D} }{E}

Find A B C D E + 1 ABCDE+1

Details and Assumptions

1) A , B , C , D , E A,B,C,D,E are positive integers. Also A B C D E ABCDE means product of the integers A , B , C , D , E A,B,C,D,E . Also base of l o g log is e e

2)Remember D D is not divisible by a perfect power(power >1 ) of any integer.

3) ζ ( s ) = n = 1 1 n s \displaystyle \zeta{(s)} = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ s } } } .


The answer is 1729.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Rajdeep Dhingra
Feb 24, 2015

Wow rajdeep you were so young when you knew all these things.....Good job...The main stake of the problem was the use of l n s i n x = σ C o s 2 n x / n l n 2 lnsinx=-\sigma Cos2nx/n-ln2 . n n varying from 1 i n f i n i t e 1-infinite .And next integration by multiplying x 2 x^2 and changing the order of summation and integration.My method was short as I used this and a prev proven x l n s i n x = 7 / 16 ζ ( 3 ) π 3 / 8 l n 2 \int xlnsinx=7/16\zeta (3)-π^3/8ln2 with the limits from 0 π / 2 0-π/2 .

Spandan Senapati - 4 years, 1 month ago
Ronak Agarwal
Feb 18, 2015

The answer is :

= π 4 ζ ( 3 ) + π 3 log 2 24 = \dfrac{\pi}{4} \zeta{(3)} + \dfrac{{\pi}^{3} \log{2} }{24}

Can you explain how one can arrive at this answer?

Calvin Lin Staff - 6 years, 3 months ago

Log in to reply

My method to this problem is just too long and I am too lazy(sorry to say like this) to post a solution to this problem. Also I posted the answer for people to confirm it who got it wrong in their attempts.

Ronak Agarwal - 6 years, 3 months ago

I will hopefully write it in 2 - 3 hours.

Rajdeep Dhingra - 6 years, 3 months ago

I have Posted the solutions in Images. Kindly edit it into Latex.

Please @Calvin Lin and sorry for @mentioning you.

Rajdeep Dhingra - 6 years, 3 months ago

Log in to reply

You have used essentially same approach as mine.

Ronak Agarwal - 6 years, 3 months ago

Log in to reply

@Ronak Agarwal Are you aware of any other ?

Rajdeep Dhingra - 6 years, 3 months ago

@Ronak Agarwal Check This Out ! here

Rajdeep Dhingra - 6 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...