Does $\int_0^1 \frac{1}{f(x)} \, dx$ converge? If $f(x)$ is a continuous function, then if $\int_1^\infty f(x)\,dx$ converges, does $\int_0^1\frac 1 {f(x)} \, dx$ converge? I answer is yes, because $\frac{1}{x^2}$ converges, but $1\frac{1}{\frac{1}{x^2}}$ also converges. in the range $[0,1].$ And in general its true.
 A: The function $f(x)\equiv 0$ is continuous, $\int_0^\infty f(x)\,dx$ converges, but $\frac{1}{f(x)}$ is not even defined.
Edit:
After the edits done to the question:
$\displaystyle \int_1^\infty f(x)\,dx$ and $\displaystyle\int_0^1\frac{1}{f(x)}\,dx$ evaluates $f$ at different points with the only exception of $x=1$. The values of $f$ at $(1,\infty)$ in no way can influence the behavior of $f$ in the interval $[0,1)$, the only condition is that the function must glue well at $1$. The function
$$f(x) = \begin{cases}
x & x\in [0,1]\\
1/x^2 & x \in [1,\infty)
\end{cases}$$
is continuous, $\displaystyle \int_1^\infty f(x)\,dx$ converges but $\displaystyle\int_0^1\frac{1}{f(x)}\,dx$ diverges.
This function is not strictly greater than $0$ because $f(0) = 0$.
Another edit:
If you want the additional requirement that $f(x)> 0$ for all $x$ then $\frac{1}{f(x)} > 0$ and is continuous in $[0,1]$. Then $\int_0^1\frac{1}{f(x)}\,dx$ converges just because $\frac{1}{f(x)}$ is a continuous function in $[0,1]$, it has nothing to do with the convergence of $\int_1^\infty f(x)\,dx$.
A: No, consider the functon $f(x)$ that's $x$ in $[0,1]$, $e^{1-x}$ everywhere else.
Another simple one: $f(x) = \exp(-x - 1/x)$.
