Let $0<a<b$. Calculate $\int_{(0,1)}\frac{t^b-t^a}{\ln(t)}dt$.

Question

Assignment:

Let $0<a<b$. Calculate $$\int_{(0,1)}\frac{t^b-t^a}{\ln(t)}dt$$

I'd appreciate a little help with this one. A hint says that rewriting $t^b-t^a$ as an integral should help, but I don't see how.

Answer 1

By the fundamental theorem of calculus, $F(b) - F(a) = \int_a^b F'(x)\,dx$. Take $F(x) = t^x$. Then consider changing the order of integration (Fubini's theorem).

Answer 2

Note that $t^b-t^a=\exp(b\log(t))-\exp(a\log(t))$ for fixed $t\in(0,1)$. Hence: $$t^b-t^a=\int_a^b \log(t)\exp(x\log(t))dx$$ Thus: $$\int_{(0,1)}\frac{t^b-t^a}{\log(t)}dt=\int_{(0,1)}\int_{(a,b)}\exp(x\log(t))dxdt$$ I think, this is computable using Tonelli's Theorem.

Answer 3

What they mean is: $$\int_{x=a}^{b}{\ln{t}*t^{x}dx}=t^x\vert_a^b = t^b-t^a$$

Is that enough to get you started?