Another integral problem

**James234** · January 11th 2007, 01:29 PM

I have:

I(a)= the integral from 0 to 1 of (x^(a)-x^(b))/(ln(x)) where a>b>-1.

How would I show that I'(a)=1/(a+1)?

**CaptainBlack** · January 11th 2007, 08:41 PM

Originally Posted by James234

I have:

I(a)= the integral from 0 to 1 of (x^(a)-x^(b))/(ln(x)) where a>b>-1.

How would I show that I'(a)=1/(a+1)?

We have:

$ I(a)= \int_0^1 \frac{x^a-x^b}{\ln(x)}dx= \int_0^1 \frac{x^a}{\ln(x)}dx- \int_0^1 \frac{x^b}{\ln(x)}dx $

so:

$ \frac{d}{da}I(a)= \frac{d}{da}\int_0^1 \frac{x^a}{\ln(x)}dx= \frac{d}{da}\int_0^1 \frac{e^{a \ln (x)}}{\ln(x)}dx $

Now differentiating under the integral gives:

$ \frac{d}{da}I(a)= \int_0^1 \ln(x) \frac{e^{a \ln (x)}}{\ln(x)}dx=\int_0^1 x^a dx = \left[ \frac{x^{a+1}}{a+1} \right]_0^1=\frac{1}{a+1} $

RonL

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