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Beta and Gamma related question?
integrate 0 to 1: (x^a -1)/(log (x)) dx
where a is a constant and a> -1
and log is base 10 log
and log is base e log not base 10
1 Answer
- kbLv 77 years agoFavorite Answer
Let g(a) = ∫(x = 0 to 1) (x^a - 1) dx/ln x.
Differentiate both sides with respect to a:
g'(a) = ∫(x = 0 to 1) (x^a ln x - 0) dx/ln x, differentiating under the integral sign
........= ∫(x = 0 to 1) x^a dx
........= x^(a+1)/(a+1) {for x = 0 to 1}
........= 1/(a+1).
Since g'(a) = 1/(a+1), integrating yields g(a) = ln(a+1) + C, since a > -1.
To solve for C, note that when a = 0, we have g(a) = ∫(x = 0 to 1) (1 - 1) dx/ln x = 0.
So, letting a = 0 yields 0 = ln(0+1) + C ==> C = 0.
Therefore,
∫(x = 0 to 1) (x^a - 1) dx/ln x = ln(a+1).
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I hope this helps!