Decent Integral (1)
Evaluate ∫ 0 1 tan − 1 ( x ) ln ( x ) d x . If the answer can be expressed as c a ln ( 2 ) + ( π − b ) π , where a , b and c are positive integers, write your answer as a + b + c .
Challenge : Solve the problem by hand.
The answer is 84.
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1 solution
I commented on Mark Hennings's solution to another problem like this (that uses a differentiation/integration under the integral sign technique), and I made the same mistake. But I can't find it to tell him I understand it now. It was integral of (ln(x))^2/e^x. Weird.
Very nice solution except I believe you made a mistake. But before I get to that, the fact that you use n in two different places makes it confusing. However, if one can recognize that the n in the series is a bound (or dummy) variable, then they can see through that part. However, you double differentiated with respect to n. Shouldn't you have integrated it w.r.t to n instead of differentiating it again? So then isn't your answer incorrect?
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i think you are confused in these two steps: ∫ 0 1 a r c t a n ( x ) ln ( x ) d x = ∫ 0 1 n = 0 ∑ ∞ 2 n + 1 x 2 n + 1 ( − 1 ) n ln ( x ) d x = n = 0 ∑ ∞ 2 n + 1 ( − 1 ) n ∫ 0 1 x 2 n + 1 ln ( x ) and d n d ∫ 0 1 x 2 n + 1 d x = ∫ 0 1 d n d x 2 n + 1 d x = 2 ∫ 0 1 x 2 n + 1 ln ( x ) d x
the way to solve this is to write the sum n = 0 ∑ ∞ 2 n + 1 ( − 1 ) n ∫ 0 1 x 2 n + 1 ln ( x ) d x = n = 0 ∑ ∞ 2 ( 2 n + 1 ) ( − 1 ) n d n d ∫ 0 1 x 2 n + 1 d x = n = 0 ∑ ∞ 2 ( 2 n + 1 ) ( − 1 ) n d n d 2 n + 2 1 = − n = 0 ∑ ∞ ( 2 n + 1 ) ( 2 n + 2 ) 2 ( − 1 ) n note the exchange of sum and integral is possible due to to absloute convergence for the series of arctan(x) between 0 and 1. we can use partial fractions to get the sum is n = 0 ∑ ∞ ( 4 ( n + 1 ) 2 ( − 1 ) n + 2 ( n + 1 ) ( − 1 ) n − 2 n + 1 ( − 1 ) n ) = 4 η ( 2 ) + 2 η ( 1 ) − arctan ( 1 ) where η ( n ) is the dirichilet eta function. the solution follows as 4 8 π ( π − 1 2 ) + 2 4 ln ( 2 )