Asymptotics of integral $ \int_{\delta}^{1}\frac{1}{\sqrt{y+e^{-y}-1}}~dy $

Question

I am interested in understanding the behaviour of the following integral $$ \int_{\delta}^{1}\frac{1}{\sqrt{y+e^{-y}-1}}~dy $$ in the limit $\delta\to0$. It arises when trying to solve a particular non-linear diffusion model for charges. Upon Taylor expanding the exponential to quadratic terms, the integrand appears to be $O(1/y)$ as $y\to0$ meaning that the integral is logarithmically singular as $\delta\to0$. However, I'm not sure how to find the precise leading order behaviour down to $O(1)$, as I can't do the integral analytically and I can't see how to use an expansion of the integrand valid across the whole region of integration. Any help in doing so is greatly appreciated.

Answer 1

If you also want to find terms in $\delta$, then you can expand the integrand as a power series. I think the easiest way to do this is honestly just to use Wolfram Alpha as shown here. Integrating term by term then gives the following series for the integral

$$\int_{\delta}^{1}{\frac{1}{\sqrt{y+e^{-y}-1}}~dy} = \left[\sqrt{2}\ln(y)+\frac{\sqrt{2}}{6}y-\frac{\sqrt{2}}{1080}y^2+\frac{\sqrt{2}}{12960}y^3-\dots\right]_{\delta}^1$$

$$= \sqrt{2}\ln\left(\frac{1}{\delta}\right)+ C -\frac{\sqrt{2}}{6}\delta+\frac{\sqrt{2}}{1080}{\delta}^2-\frac{\sqrt{2}}{12960}{\delta}^3+\dots$$

where $C$ is the sum of all the coefficients, namely

$$\sqrt{2}\left(\frac{1}{6}-\frac{1}{1080}+\frac{1}{12960}-\frac{1}{181440}+\frac{1}{680400}-\dots\right) = 0.1658138 \text{ to the first $7$ d.p}$$

To find the exact expansion, there is a well-known series for $\frac{1}{\sqrt{1+x}}$ when $|x|<1$, so it makes sense to first put the integrand into this form. Notice that

$$y+e^{-y}-1 = -1+y+\sum_{n=0}^{\infty}{\frac{(-1)^n}{n!}y^n}$$

$$= \sum_{n=2}^{\infty}{\frac{(-1)^n}{n!}y^n} = \frac{y^2}{2} - \frac{y^3}{6} + \dots$$

Pulling out the leading term $y^2/2$, we obtain

$$y+e^{-y}-1 = \frac{y^2}{2}\left(1 - \sum_{n=3}^{\infty}{\frac{2(-1)^{n-1}}{n!}y^{n-2}}\right)$$

where the inside is of the form $1+x$ with $|x|<1$. So plugging this into the integral, we get

$$\int_{\delta}^{1}{\frac{1}{\sqrt{y+e^{-y}-1}}~dy} = \int_{\delta}^{1}\frac{\sqrt{2}}{y\sqrt{1 - \sum_{n=3}^{\infty}{\frac{2(-1)^{n-1}}{n!}y^{n-2}}}}~dy$$

$$= \sqrt{2}\int_{\delta}^1{\left(\sum_{m=0}^{\infty}{\binom{-1/2}m(-1)^m\left(\sum_{n=3}^{\infty}{\frac{2(-1)^{n-1}}{n!}y^{n-2}}\right)^m}\right)\frac{dy}{y}}$$

$$= \sqrt{2}\sum_{m=0}^{\infty}{\binom{-1/2}m(-1)^m\int_{\delta}^1{\left(\sum_{n=3}^{\infty}{\frac{2(-1)^{n-1}}{n!}y^{n-2}}\right)^m}\frac{dy}{y}}$$

$$= \sqrt{2}\sum_{m=0}^{\infty}{\binom{-1/2}m(-1)^m\int_{\delta}^1{\left(\frac{y}{3}-\frac{y^2}{12}+\frac{y^3}{60}-\dots\right)^m}\frac{dy}{y}}$$

Though way more complicated, this mess when expanded and integrated should give you the same coefficients as those shown above.

Answer 2

$$e^{-y}=1-y+\frac12y^2-\mathcal O(y^3)$$

hence

$$\sqrt{y+e^{-y}-1}=\sqrt{\frac12y^2-\mathcal O(y^3)}=\frac1{\sqrt2}y-\mathcal O(y^2)$$

The divergent part of the integral is approximated as

$$ \int_{\delta}^{1}\frac{dy}{\sqrt{y+e^{-y}-1}}\sim \int_{\delta}^{1}\frac{1}{\frac1{\sqrt2}y}~dy $$

hence $$ \int_{\delta}^{1}\frac{dy}{\sqrt{y+e^{-y}-1}}- \int_{\delta}^{1}\frac{\sqrt2}{y}~dy=\text{Constant} $$

and we get

$$\int_{\delta}^{1}\frac{dy}{\sqrt{y+e^{-y}-1}}=-\sqrt2\cdot \ln\delta+C+\mathcal o(1)$$

Answer 3

To get quite accurate approximations consider the simple $[2n,2n+1]$ Padé approximant $P_n$ of the integrand around $y=0$ (you can easily build it from the Taylor series).

A simple one could be $P_1$ which is $$\frac{1}{\sqrt{y+e^{-y}-1}}\sim \frac { \sqrt{2}+\frac{1}{2 \sqrt{2}}y+\frac{7 }{180 \sqrt{2}}y^2} {y+\frac{1}{12}y^2+\frac{1}{180}y^3 }=\frac{\sqrt{2}}{y}+\frac{5 (y+12)}{\sqrt{2} \left(y^2+15 y+180\right)}$$ whose error is $\sim 10^{-5}y^4$

Nothing very complicated and, for the definite integral, $$\int_\delta^1\frac{dy}{\sqrt{y+e^{-y}-1}}=-\sqrt{2} \log (\delta )+3 \sqrt{\frac{5}{22}} \tan ^{-1}\left(\frac{3 \sqrt{55} (1-\delta )}{375+17 \delta}\right)+$$ $$\frac{5}{2 \sqrt{2}}\log \left(\frac{196}{180+15 \delta +\delta ^2}\right)$$ This seems to be quite decent even for "large" values of $\delta$. $$\left( \begin{array}{ccc} \delta & \text{approximation} & \text{numerical integration} \\ 10^{- 1} & 3.468069897 & 3.468071552 \\ 10^{- 2} & 6.745629734 & 6.745631389 \\ 10^{- 3} & 10.00409812 & 10.00409978 \\ 10^{- 4} & 13.26065732 & 13.26065897 \\ 10^{- 5} & 16.51702560 & 16.51702725 \\ 10^{- 6} & 19.77337479 & 19.77337644 \\ 10^{- 7} & 23.02972207 & 23.02972372 \\ 10^{- 8} & 26.28606916 & 26.28607081 \\ 10^{- 9} & 29.54241623 & 29.54241788 \\ 10^{- 10} & 32.79876329 & 32.79876495 \\ \end{array} \right)$$

For sure, the next one would be much better (for $\delta=10^{-1}$, the result would be $3.468071555$) but the formula would not be very nice since involving the roots of cubic polynomials.