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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!

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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 $$\int_a^1 t^{-\frac{1}{2}}f(t)\mathrm{d}t\leq M\int_a^1t^{-\frac{1}{2}}\mathrm{d}t=2M[t^{\frac{1}{2}}]_{t=a}^{t=1}=2M(1-\sqrt{a})$$

By the monotonicity of the integral. Taking the limit yields the result.

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  • $\begingroup$ I didn't modify my answer in light of Shashi more direct answer, but since you accepted mine, I'll add that the limit is unnecessary and you can work directly on the integral like Shahi did. $\endgroup$ – user438666 Nov 12 '17 at 20:29
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$f$ 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.

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