Test the improper integral ∫ e-x1/2 / x1/2 for convergence

Test the following improper integral for convergence:

$\int_{0^+}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx.$

The integral converges.

Proof. We can compute this integral directly. First, we evaluate the indefinite integral using the substitution $u = \sqrt{x}$ , $du = \frac{1}{2 \sqrt{x}} \, dx$ .

$\begin{align*} \int \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx &= 2 \int e^{-u} \, du \\[9pt] &= -2 e^{-u} \\[9pt] &= -2 e^{-\sqrt{x}} \, dx. \end{align*}$

Now, we have discontinuities at both limits of integration so we evaluate by taking two limits,

$\begin{align*} \int_{0^+}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx &= \lim_{\substack{b \to +\infty \\ a \to 0^+}} \int_a^b \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx \\[9pt] &= \lim_{\substack{b \to +\infty \\ a \to 0^+}} \left( -2 e^{-\sqrt{x}} \Bigr \rvert_a^b \right) \\[9pt] &= \lim_{\substack{b \to +\infty \\ a \to 0^+}} \left( -2 e^{-\sqrt{b}} + 2 e^{-\sqrt{a}} \right) \\[9pt] &= 2. \qquad \blacksquare \end{align*}$

Stumbling Robot

Test the improper integral ∫ e^{-x^1/2} / x^1/2 for convergence

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