I need to prove that $$\int_{1}^{b}a^{\frac{\log x}{\log b}} dx > \ln b$$
I know that if somehow i can prove the function to be greater than $1/x$ then i would be able to prove the rest. but i can`t think of anything further.
Any help will be appreciated.
You can compute the integral: $$a^\frac{\ln x}{\ln b}=\exp \left(\frac{\ln x \ln a}{\ln b} \right)=x^\frac{\ln a}{\ln b}$$ so if $\frac{\ln a}{\ln b} \neq -1$ : $$I=\int_1^ba^\frac{\ln x}{\ln b} dx=\left[\frac{1}{\frac{\ln a}{\ln b} +1}x^{\left(\frac{\ln a}{\ln b} +1\right)}\right]_1^b$$ once more using the properties of the $\ln$: $$b^{\left(\frac{\ln a}{\ln b} +1\right)}=\exp\left( \ln b\left(\frac{\ln a}{\ln b} +1 \right)\right)=\exp(\ln a +\ln b)=ab$$ so: $$I=\ln(b) \left(\frac{\exp(\ln a +\ln b)-1}{\ln(a)+\ln(b)}\right)$$
From there if $\ln b$ and $\ln b+\ln a$ have the same sign you can use that $\forall z \in \Bbb R$ $e^z \geq 1+z$ to show that:
If $\ln(b) < 0$ and $\ln(b)+\ln(a) < 0$: $$\frac{\exp(\ln a +\ln b)-1}{\ln(a)+\ln(b)} < 1$$ so $$I > \ln(b)$$
In the other cases the inequality seems false.