20080306, 03:10  #1 
8,779 Posts 
Impossible Integral
Alright, so this is a Calculus 2 homework problem and my friends and I are completely stuck.
The original question is to find if the Summat of {tan^1 (x)}/x^1.1 (from 1 to infinity)converges or diverges, and if it converges to find its sum. Using the Integral Test, we said that this series, did in fact converge. The problem was in evaluating the integral. Integral (1 to infinity) of (tan^1 (x))/x^1.1 We cannot manage to evaluate this... no theorem or rule we can find helps in this evaluation. Any help would be greatly appreciated. Thanks! 
20080306, 15:46  #2  
Nov 2003
2^{6}·113 Posts 
Quote:
simply note that the numerator is bounded above by pi, and that sum(pi/x^1.1)) clearly converges. (2) Evaluating the integral that you gave will not give the sum of the series. For that, you need EulerMacLauren summation. (3) The indefinite integral can probably be expressed in terms of hyper geometric functions. It is very unlikely that a closed form expression exists for the sum. (4) To evaluate your infinite integral, I would use complex contour integration methods, along with Cauchy's Theorem. Start by finding the residues of the series; this requires a Laurent expansion. I won't even attempt it by hand. I would want to use Mathematica. I would use a quartercircle contour centered at 0. The integral can't be expressed in terms of elementary functions. Finding the sum seems well beyond a 2nd year calculus course. 

20080306, 21:16  #3 
5·577 Posts 
The original problem was to find if SIGMA [(tan^1 (n)/n^1.1)] from 1 to infinity converges and if it converges to find its sum. I proved that it converges using the Integral test. Using 2nd year calculus tools, how would you find the sum? A 3rd year calc student mentioned something about a comparison that could be done involving pi/2.

20080307, 12:30  #4  
Nov 2003
2^{6}×113 Posts 
Quote:
It *might* be expressible in terms of hypergeometric functions. Why do I think this? Because integral ATAN(x)/x dx has such a representation. 

20080307, 12:42  #5 
"Robert Gerbicz"
Oct 2005
Hungary
1410_{10} Posts 
By Mathematica 5.1 the integral is (see the attachment)

20080308, 20:05  #6 
Feb 2007
1B0_{16} Posts 
since convergence is indeed ensured by comparision with (pi/2)/x^1.1,
why not do it numerically... gp> intnum(x=1,[1],atan(x)/x^1.1) %1 = 14.879905283440983234832458288559145866892862726910 gp> ## *** last result computed in 140 ms. But here's a primitive from Maple: 10/x^(1/10)*arctan(x) +5/2*sqrt(2)*ln((x^(1/5)+x^(1/10)*sqrt(2)+1)/(x^(1/5)x^(1/10)*sqrt(2)+1)) +5*sqrt(2)*arctan(x^(1/10)*sqrt(2)+1) +5*sqrt(2)*arctan(x^(1/10)*sqrt(2)1) 20*sum(_R*ln(x^(1/10)262144*_R^9),_R = RootOf(4294967296*_Z^1616777216*_Z^12+65536*_Z^8256*_Z^4+1)) Taking the limit x=infinity seems obvious except for the last term. (might have changed 4Z into Z'...) 
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