I need to find if $$\int_0^\infty \frac {e^{-x}-1} {\sqrt x^3}dx$$ converges.

Let $$f(x) = \int_0^\infty \frac {1- e^{-x}} {\sqrt x^3}dx \ge 0$$

$$\int_0^\infty \frac {1-e^{-x}} {\sqrt x^3}dx = \int_0^1 \frac {1- e^{-x}} {\sqrt x^3}dx + \int_1^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$$

The integral $$\int_1^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$$ converges by applying the comparison test: $$\lim_{x \to \infty}\frac {\frac {1-e^{-x}} {\sqrt {x^3}}} {\frac 1 {\sqrt {x^3}}} = \lim_{x \to \infty}1- e^{-x}=1$$

so $f(x)$ converges iff $$\int_1^\infty \frac 1 {x^{\frac 3 2}}dx$$


the problem i have is with $\int_0^1 f(x)dx$

i tried to use the same comparison test but got limit 0, so divergence of $$\int_0^1 \frac 1 {x^{\frac 3 2}}dx$$ doesn't indicate that $\int_0^1 f(x)dx$ diverges.

how do i handle $\int_0^1 f(x)dx$ ?

  • $\begingroup$ Note that $1-e^{-x}$ behaves as $x$ for small $x$ so there is no problem with the integrand as $x$ tends to $0$. $\endgroup$
    – Urgje
    Jul 10, 2014 at 21:47
  • $\begingroup$ If you expand $e^{-x}$, you see $$1-e^{-x} = x - \frac{x^2}{2} + \frac{x^3}{6} - \dotsc$$ $\endgroup$ Jul 10, 2014 at 21:48
  • $\begingroup$ Use subsitution $x = y^2$. See my post... $\endgroup$ Jul 10, 2014 at 21:59

3 Answers 3


Since $\int_{0}^{1}\frac{1}{x^{1/2}}$ converges and $\displaystyle\lim_{x\to 0^{+}}\frac{\frac{1-e^{-x}}{x^{3/2}}}{\frac{1}{x^{1/2}}}=\lim_{x\to 0^{+}}\frac{1-e^{-x}}{x}=\lim_{x\to 0^{+}}\frac{e^{-x}}{1}=1$,

$\;\;\;\;\;\int_{0}^{1}\frac{1-e^{-x}}{x^{3/2}}$ converges.


This integral is not too difficult to just go ahead and solve completely. Starting by integrating by parts,

$$\begin{align} \int_{0}^{\infty}\frac{e^{-x}-1}{x^{3/2}}\mathrm{d}x &=\left[-\frac{2}{x^{1/2}}(e^{-x}-1)\right]_{0}^{\infty}-\int_{0}^{\infty}\frac{2\,e^{-x}}{x^{1/2}}\mathrm{d}x\\ &=0-0-2\int_{0}^{\infty}x^{\frac12-1}e^{-x}\mathrm{d}x\\ &=-2\,\Gamma{\left(\frac12\right)}\\ &=-2\,\sqrt{\pi}, \end{align}$$

where we've substituted the definition of the Gamma function and used the fact that $\Gamma{\left(\frac12\right)}=\sqrt{\pi}$. Since the integral converges to the limit $-2\,\sqrt{\pi}$, the integral obviously converges.


Suppose you didn't know the value of $\Gamma{\left(\frac12\right)}$. An easy way to establish that it converges is to make a substitution that transforms the it into the Gaussian integral, whose convergence is probably common knowledge even for most beginning students. Substituting $x=u^2$, we have $u=x^{1/2}$ and $\mathrm{d}u=\frac{\mathrm{d}x}{2\,x^{1/2}}$, and hence:

$$\begin{align} \Gamma{\left(\frac12\right)} &=\int_{0}^{\infty}x^{\frac12-1}e^{-x}\mathrm{d}x\\ &=2\int_{0}^{\infty}e^{-x}\frac{\mathrm{d}x}{2\,x^{\frac12}}\\ &=2\int_{0}^{\infty}e^{-u^2}\mathrm{d}u. \end{align}$$



$$ \int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx. $$

Use substitution $x = y^2$, then we obtain

$$ \begin{eqnarray} \int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx &=& 2 \int_0^\infty \frac{ \exp(-y^2) - 1 }{ y^2 } dy\\ &=& \color{red}{\left[ - 2 \frac{ \exp(-y^2) - 1 }{y} \right]_0^\infty} - \color{blue}{\int_0^\infty \exp(-y^2) dy}\\ &=& \underbrace{\color{red}{- 2 \lim_{y \rightarrow 0} \frac{ \exp(-y^2) - 1 }{y}}}_{\textit{use l'Hôpital}} - \color{blue}{2 \sqrt{\pi}}\\ &=& \color{red}{4 \lim_{y \rightarrow 0} \exp(-y^2)} - \color{blue}{2 \sqrt{\pi}}\\ &=& \color{blue}{- 2 \sqrt{\pi}}. \end{eqnarray} $$


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