# Improper Integrals and general continuous functions $\int^1_0t^{-\frac{1}{2}}f(t)dt$

Let $f:[0,1]\to\mathbb{R}$ be continuous, then i want to show that the integral $$\int^1_0t^{-\frac{1}{2}}f(t)dt$$ is convergent.

I know I need to use the improper integral $$\lim_{a\to0}\int^1_at^{-\frac{1}{2}}f(t)dt$$ I have been working through some similar improper integral problems and this seems fairly simple but I can't seem to get right, I feel that the continuity isn't a strong enough condition on f. I have tried by parts and using a method of that similar to proving the gamma functions convergence but I can't get anything useful out. Please Help, thanks!

If $f$ is continuous on $[0,1]$, then $f$ has a maximum in the interval. Let $M=\max_{x\in[0,1]}f$. So you have that $$\int_a^1 t^{-\frac{1}{2}}f(t)\mathrm{d}t\leq M\int_a^1t^{-\frac{1}{2}}\mathrm{d}t=2M[t^{\frac{1}{2}}]_{t=a}^{t=1}=2M(1-\sqrt{a})$$
$f$ continuous on $[0,1]$ so by Weierstrass Theorem there is a $M$ such that $|f|\leq M$. So: \begin{align} \int^1_0 |t^{-1/2} f(t) | dt\leq M\int^1_0 t^{-1/2} dt \end{align} You make the conclusion.