I = ∫ 0 π / 2 x 2 lo g ( sec ( x ) ) d x
If the value of I can be represented as = A π ζ ( B ) + E π C lo g D
Find A B C D E + 1
Details and Assumptions
1) A , B , C , D , E are positive integers. Also A B C D E means product of the integers A , B , C , D , E . Also base of l o g is e
2)Remember D is not divisible by a perfect power(power >1 ) of any integer.
3) ζ ( s ) = n = 1 ∑ ∞ n s 1 .
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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 . n varying from 1 − i n f i n i t e .And next integration by multiplying 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 / 1 6 ζ ( 3 ) − π 3 / 8 l n 2 with the limits from 0 − π / 2 .
The answer is :
= 4 π ζ ( 3 ) + 2 4 π 3 lo g 2
Can you explain how one can arrive at this answer?
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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.
I will hopefully write it in 2 - 3 hours.
I have Posted the solutions in Images. Kindly edit it into Latex.
Please @Calvin Lin and sorry for @mentioning you.
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You have used essentially same approach as mine.
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@Ronak Agarwal – Are you aware of any other ?
@Ronak Agarwal – Check This Out ! here
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