Use the comparison Theorem to determine whether the integral is convergent or divergent
integral from 0,1
[e^(-x)]/sqrt(x)
thanks guys
$\displaystyle \frac{e^{-x}}{\sqrt{x}} \leq \frac{e^{\sqrt{x}}}{\sqrt{x}} \,\ \forall \,\ x \in (0,1]$.
and it is easy to show that the improper integral $\displaystyle \int_0^1 \frac{e^{\sqrt{x}}}{\sqrt{x}} \,\ dx$ converges.
Hence, your improper integral converges.
T
Similar to what Krizalid has shown, we can also find a lower bound.
For all $\displaystyle x\in[0,1]$, we have
$\displaystyle \frac{1}{\mathrm{e}}\leq\mathrm{e}^{-x}\leq1$, which yields by integrating after dividing by $\displaystyle \sqrt{x}$
$\displaystyle \frac{2}{\mathrm{e}}\leq\int_{0}^{1}\frac{\mathrm{ e}^{-x}}{\sqrt{x}}\mathrm{d}x\leq2.$
Note that above, we apply improper integration and that
$\displaystyle \int_{a}^{b}\frac{1}{\sqrt{x}}\mathrm{d}x=2\sqrt{x }\big|_{x=a}^{x=b}$ provided that $\displaystyle b>a>0$.