Science:Math Exam Resources/Courses/MATH101/April 2016/Question 05 (a)/Solution 1

Set $f(x)={\frac {e^{-{\sqrt {x}}}}{\sqrt {x}}}$.

For $x\geq 1$, this function is positive and decreasing (since it is the product of the two decreasing functions $e^{-{\sqrt {x}}}$ and ${\frac {1}{\sqrt {x}}}$).

Using the substitution $u={\sqrt {x}}$, so that $du={\frac {1}{2{\sqrt {x}}}}\,dx$, we see that

${\begin{aligned}\int _{1}^{\infty }f(x)dx&=\lim _{R\to \infty }\int _{1}^{R}\left({\frac {e^{-{\sqrt {x}}}}{\sqrt {x}}}dx\right)\\&=\lim _{R\to \infty }\left(\int _{1}^{\sqrt {R}}e^{-u}\cdot 2\,du\right)\\&=\lim _{R\to \infty }\left({-}2e^{-u}{\Big |}_{1}^{\sqrt {R}}\right)\\&=\lim _{R\to \infty }\left({-}2e^{-{\sqrt {R}}}+2e^{-1}\right)=0+2e^{-1},\end{aligned}}$

and so this improper integral converges. By the Integral Test, the given series also converges.