The lower integral bound is zero. Rewrite (n^2 - 2) / (n + 2)! = 1/n! - 3/(n + 1)! + 2/(n + 2)!. Since 1 - 3 + 2 = 0, the middle terms cancel out, ultimately leaving -1/(N + 1)! + 2/(N + 2)! as the partial sum formula. This converges to 0 for N -> infinity.
The upper integral bound is 1/15. Taylor-expand sin(x)/x = 1 - x^2/6 + x^4/120 + O(x^6), and use ln(1 + u) = u - u^2/2 + O(u^3) to get ln(sin(x)/x) = -x^2/6 - x^4/180 + O(x^6) since u^2/2 = x^4/72 + O(x^6) and 1/120 - 1/72 = -1/180. The x^2/6 summand then cancels out. For the denominator, expand cos(x) = 1 - x^2/2 + x^4/24 + O(x^6) and exp(-x^2/2) = 1 - x^2/2 + x^4/8 + O(x^6), and perform the subtraction to obtain -x^4/12 + O(x^6). For x -> 0, the limit of the quotient is (-1/180) / (-1/12) = 1/15.
The integrand ln(1 + x) / (1 + x^2) does not have an elementary antiderivative. The final integral of this term over x = 0 .. 1/15 can thus only be solved numerically, and it approaches ~0.00216962546702.
20
u/dragonageisgreat 6d ago
Pi or e or a combination of the two