Is the integrand integrable?

Question

Suppose $f(\tau)$ is an integrable function on $[0,t]$. Also assuming that $\alpha$ be a real and positive number. Can it be proved that the integrand is still integrable $\int_{0}^{t} \frac{f(\tau)}{(t-\tau)^{-\alpha+1}} \,d\tau$ ?

Answer 1

Take $$f\left(\tau\right)=\ln\left(1+\tau\right)$$ With $\alpha=1$ $$ \int_{0}^{t}\frac{\ln\left(1+\tau\right)}{\left(t-\tau\right)^{2}}\text{d}\tau \text{ does not exist} $$

EDIT : $$ \frac{f\left(\tau\right)}{\left(t-\tau\right)^{1-\alpha}} \underset{(0^{+})}{\sim}\frac{f\left(\tau\right)}{t^{1-\alpha}} $$ If $f$ is continuous on $\left[0,t\right]$ then $$ \frac{f\left(\tau\right)}{\left(t-\tau\right)^{1-\alpha}} \underset{(0^{+})}{\sim}\frac{f\left(0\right)}{t^{1-\alpha}} $$ And $\displaystyle t \mapsto \frac{1}{t^{1-\alpha}}$ is integrable on $\left[0,1\right[$