Let $f(x) = \int^\sqrt{x}_1 e^{-t^2}dt$. Find $\int^1_0 \frac{f(x)}{ \sqrt{x}}dx$.

Could anyone give me any hint how to start? The Erf function $f(x)$ seems not to be easily integrated.

  • $\begingroup$ There is a document titled "A table of integrals of the error functions" by edward W. Ng and Murrey Geller. I think that could be helpful to you. $\endgroup$ – Frank Moses Nov 8 '16 at 0:56
  • $\begingroup$ The answer, as a side note, is $\frac{1}{e}-1$ according to 1 minute of using Wolfram Alpha $\endgroup$ – Brevan Ellefsen Nov 8 '16 at 0:57
  • $\begingroup$ @FrankMoses This is a test question so I could not possibly refer to any docimentation. $\endgroup$ – user122049 Nov 8 '16 at 0:57
  • 1
    $\begingroup$ Try integration by parts and apply fundamental theorem of calculus. $\endgroup$ – ℵ_ϵ Nov 8 '16 at 0:57
  • $\begingroup$ As Brevan Ellefsen found, the result is $1/e -1$, according to Mathematica. $\endgroup$ – David G. Stork Nov 8 '16 at 0:58

You want to find
This should SCREAM integration by parts, as we somehow want to differentiate the numerator. Let's try it. We note that $u' = \frac{e^{-x}}{2\sqrt{x}}$ and $v=2\sqrt{x}$ $$=2\sqrt{x}f(x)\bigg|_0^1 -\int_0^1 \frac{e^{-x}2\sqrt{x}}{2\sqrt{x}}$$ $$=2f(1) -\int_0^1 \frac{e^{-x}2\sqrt{x}}{2\sqrt{x}}$$ $$=\left(2\int_1^{1}e^{-t^2}dt\right) -\int_0^1 e^{-x}$$ $$=\frac{1}{e}-1$$

  • $\begingroup$ All too easy .... +1 $\endgroup$ – Mark Viola Nov 8 '16 at 2:39

Brevan Ellefsen's solution is the most straightforward and efficient. Here is an alternative solution.

\begin{equation} f(x) = \int\limits_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt \label{eq:161108-1} \tag{1} \end{equation} \begin{equation} \int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx \label{eq:161108-2} \tag{2} \end{equation}

We need the following result \begin{equation} \int \mathrm{erf}(x) dx = x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}} \label{eq:161108-3} \tag{3} \end{equation} Proof: Integrate by parts \begin{align} \int \mathrm{erf}(x) dx &= x\,\mathrm{erf}(x) -\frac{2}{\sqrt{\pi}} \int x\,\mathrm{e}^{-x^{2}} dx \\ &= x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}} \end{align} we used the substitution $u=x^{2}$.

Now we evaluate equation \eqref{eq:161108-1} \begin{equation} f(x) = \frac{\sqrt{\pi}}{2} \mathrm{erf}(x) \Big|_{1}^{\sqrt{x}} = \frac{\sqrt{\pi}}{2} [\mathrm{erf}(\sqrt{x}) - \mathrm{erf}(1)] \label{eq:161108-4} \tag{4} \end{equation}

Substitute equation \eqref{eq:161108-4} into equation \eqref{eq:161108-2} and evaluate the following two integrals.

\begin{align} I_{1} &= \int\limits_{0}^{1} \frac{\mathrm{erf}(\sqrt{x})}{\sqrt{x}} dx \\ &= 2\int\limits_{0}^{1} \mathrm{erf}(z) dz \\ &= 2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} - \frac{2}{\sqrt{\pi}} \end{align} we used the substitution $z=\sqrt{x}$.

\begin{equation} I_{2} = \int\limits_{0}^{1} \frac{1}{\sqrt{x}} dx = 2 \end{equation}

Putting all of the pieces together yields our final result \begin{align} \int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx &= \frac{\sqrt{\pi}}{2} \left(2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} - \frac{2}{\sqrt{\pi}} - 2\,\mathrm{erf}(1) \right) \\ &= \frac{1}{\mathrm{e}} - 1 \end{align}

  • $\begingroup$ I like this answer a lot. Really shows why the integral simplifies so much. The solution I posted expresses this by-way-of the fractions all simplifying a lot, but yours is definitely the winner in regards to showing how the non-elementary parts cancel $\endgroup$ – Brevan Ellefsen Nov 8 '16 at 14:09

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.