1/alna([x])(a-1)
\int_{0}^{[x]}{a^t/a^[^t^]}dt (a>0), where [] denotes greatest int function is?
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12 Answers
Subash
·2009-03-30 03:12:52
split the integral
by taking limits 0-1 and 1-2.....[x-1]-[x]
then you can do it
tapanmast Vora
·2009-03-30 03:16:53
by using the fact that {x} is a periodic functn ........ {x} = frac. part of x
I = [x] ∫a^tdt ................ the integral is over 0 to 1
ANKIT MAHATO
·2009-03-30 03:21:00
0∫[x] at - [t] dt
= 0∫[x] a{t} dt .. now {.} is a periodic function with period 1
=[x] 0∫1 a{t} dt = [x] 0∫1 at dt
= [x] (a - 1)/ln a