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 Intriguing Integral Illation II (Posted on 2014-04-03)
Evaluate this integral in terms of a and b, where each of a and b is a positive real number.

∫ Arctan(ax)−Arctan(bx) dx
0

 No Solution Yet Submitted by K Sengupta No Rating

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 A little more direct.... Comment 2 of 2 |
The derivative of arctan(ax) is a/(a^2*x^2+1)

The indefinite integeral of arctan(ax) is x*arctan(ax) - ln(a^2*x^2+1)/(2a)

The indefinite integeral, when evaluated at 0 is 0. So the limit may be expressed as:
lim {x to inf} [x*arctan(ax) - x*arctan(bx)] - [ln(a^2*x^2+1)/(2a) - ln(b^2*x^2+1)/(2b)]

The first half of the limit can be expressed as [arctan(ax) - arctan(bx)]/[1/x], to which L'Hopital's rule may be applied.  Then the limit simplifies to 1/b - 1/a.

The second half of the limit simplifies into ln(a)/a - ln(b)/b + ln(x)*(1/a-1/b).  Then this limit diverges to infinity.

So the combined limit also diverges to infinity.

 Posted by Brian Smith on 2016-07-04 23:01:03

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