I came across this : I'm trying to evaluate it up to $ o(\epsilon) $ $$ F\left(\varepsilon\right)=\int\limits _{0}^{1}\frac{1}{\sqrt{x+\varepsilon}} \, \mathrm{d}x $$

I've trying considering to look at it as the following $$ F\left(\varepsilon\right)=\int\limits _{0}^{\theta}\frac{1}{\sqrt{x+\varepsilon}} \, \mathrm{d}x+\int\limits _{\theta}^{1}\frac{1}{\sqrt{x+\varepsilon}}\, \mathrm{d}x $$

and evaluate each integral separately and hoping that the intermediate region will cancel out (this method is widely used in perturbation theory to evaluate integrals) but I can't seem to find the right way to evaluate the first integral...

Any help?


  • $\begingroup$ My idea: it is not that hard to see that $(x+\varepsilon)^{-1/2}=x^{-1/2} -\frac{\varepsilon}{2 x^{3/2}} + o(\varepsilon x^{-3/2})$. This is a useful approximation for large $x$. So choose a splitting point $\theta$. Then approximate the integral on $[\theta,1]$ as I described above, and approximate the integral on $[0,\theta]$ using the simple fact that it is between $0$ and $2\sqrt{\theta}$. Now the part I'm not so sure about (which is why I'm not writing an answer) is how to nicely choose $\theta$ to make this a useful approximation. $\endgroup$ – Ian Jul 3 '15 at 18:49
  • $\begingroup$ (Cont.) You need $\theta$ so large that $\varepsilon \theta^{-3/2}$ is small, but you also need it so small that $2\sqrt{\theta}$ is a good estimate for the integral on $[0,\theta]$. So there is a balance to be struck. And I'm not sure how that balance is going to get the $-2\sqrt{\varepsilon}$ term from the Taylor approximation of the explicit solution. $\endgroup$ – Ian Jul 3 '15 at 18:51
  • $\begingroup$ @Ian That a good way of thinking, but as you said, the tricky part is for $ [0,\theta] $ for that region im stuck with how to expand the expression to get a solution that is allso $ \varepsilon^{2}\theta^{-\frac{3}{2}} $ $\endgroup$ – Simba Jul 3 '15 at 19:02
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    $\begingroup$ Another thought: as you can see from the explicit solution, $\int_\varepsilon^1 x^{-1/2} dx$ is a $O(\varepsilon)$ approximation. There are two pieces to its error: $\int_0^\varepsilon (x+\varepsilon)^{-1/2}$ and $\int_\varepsilon^1 (x+\varepsilon)^{-1/2}-x^{-1/2} dx$. Somehow there is some cancellation between these two errors, because the first error is on the order of $\varepsilon^{1/2}$, which means the second one must also be of the same order and with a coefficient of the opposite sign. $\endgroup$ – Ian Jul 4 '15 at 14:07
  • $\begingroup$ Thats exactly what i expect, My problem is just figuring out how to evaluate the first region integral. it cannot be expanded around x=0 ... $ \int\limits _{0}^{\theta}\frac{1}{\sqrt{x+\epsilon}}dx = \frac{1}{\sqrt{\varepsilon}}\int\limits _{0}^{\theta}1-\frac{1}{2}\frac{x}{\varepsilon}+\frac{3}{8}\left(\frac{x}{ \epsilon}\right)^{2}+O\left(\left(\frac{x}{\varepsilon}\right)^{3}\right)dx $ $\endgroup$ – Simba Jul 4 '15 at 18:01

We have that $$\int_0^1 \frac{1}{\sqrt{x+\varepsilon}} \, \mathrm{d}x = 2\left(\sqrt{\varepsilon + 1} - \sqrt{\varepsilon}\right).$$

Since we can write $1/\sqrt{x+\varepsilon} = (x+\varepsilon)^{-1/2}$ and use the reverse chain rule.

  • $\begingroup$ thanks for the advice but i need to evaluate without explicitly solving :D $\endgroup$ – Simba Jul 3 '15 at 18:21
  • $\begingroup$ @Simba You can do a series representation of the right side and get the final result that you want. In particular you have $\sqrt{\varepsilon+1}=1+\frac{\varepsilon}{2} + o(\varepsilon)$, so that the integral is $2-2\sqrt{\varepsilon}+\varepsilon+o(\varepsilon)$. $\endgroup$ – Ian Jul 3 '15 at 18:22
  • $\begingroup$ @Ian the idea is to evaluate it without solving the integral. you can expand the integrand and solve polinomial expression but not solve it directly... $\endgroup$ – Simba Jul 3 '15 at 18:22

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