Taylor series of $\int_a^b f(x,c)dx$ as $c$ approaches 0

Question

Let $p>0$ be a constant. Let \begin{equation} F(c)=\int_a^b f(x,c)dx=\int_0^1 \sqrt{x^{2p}+c^2}dx. \end{equation}

Is there any simple method to find the Taylor series of $f(c)$ at $c=0$? (At least find $k_1$ below.) \begin{equation} F(c)=\frac{1}{p+1}+k_1c+k_2c^2+\cdots \end{equation}

Thanks.

Answer 1

Let's consider the case $p>1$ first. Denoting $\,\displaystyle I(p,c)=\int_0^1\sqrt{x^{2p}+c^2}dx$, we get after straightforward transformations $$I(p,c)\overset{x^p=t}{=}\frac1p\int_0^1t^{\frac1p-1}\sqrt{t^2+c^2}dt\overset{t=cx}{=}\frac{c^{\frac1p+1}}p\int_0^\frac1cx^{\frac1p-1}\sqrt{1+x^2}dx$$ $$\overset{t=\frac1x}{=}\frac{c^{\frac1p+1}}p\int_c^\infty t^{-2-\frac1p}\sqrt{1+t^2}dt$$ Integrating by part $$=-\frac{c^{\frac1p+1}}p\frac{t^{-1-\frac1p}\sqrt{1+t^2}}{1+\frac1p}\,\bigg|_c^\infty+\frac{c^{\frac1p+1}}p\frac{1}{1+\frac1p}\int_c^\infty\frac{t^{-\frac1p}}{\sqrt{1+t^2}}dt$$ $$=\frac{\sqrt{1+c^2}}{1+p}+\frac{c^{\frac1p+1}}{1+p}\int_0^\infty\frac{t^{-\frac1p}}{\sqrt{1+t^2}}dt-\frac{c^{\frac1p+1}}{1+p}\int_0^c\frac{t^{-\frac1p}}{\sqrt{1+t^2}}dt$$ $$I(p,c)=\frac{\sqrt{1+c^2}}{1+p}+\frac{c^{\frac1p+1}}{1+p}\left(I_1-I_2\right)\tag{1}$$ where $$I_1=\int_0^\infty\frac{t^{-\frac1p}}{\sqrt{1+t^2}}dt\overset{x=\frac1{1+t^2}}{=}\frac12\int_0^1x^{\frac1{2p}-1}(1-x)^{-\frac1{2p}-\frac12}dx$$ $$=\frac12B\left(-\frac1{2p}+\frac12;\frac1{2p}\right)=\frac1{2\sqrt\pi}\Gamma\left(\frac12-\frac1{2p}\right)\Gamma\left(\frac1{2p}\right)\tag{2}$$ To evaluate $I_2$, we decompose the integrand into the series ($t\leqslant c\ll1$) $$I_2=\int_0^c\frac{t^{-\frac1p}}{\sqrt{1+t^2}}dt \sim\int_0^ct^{-\frac1p}\sum_{k=0}^\infty\binom{-\frac12}{k}t^{2k}dt=\sum_{k=0}^\infty\binom{-\frac12}{k}\frac{c^{2k+1-\frac1p}}{2k+1-\frac1p}\tag{3}$$ Putting (2) and (3) into (1) $$\boxed{\,\,I(p,c)\sim\frac{\sqrt{1+c^2}}{1+p}+\frac{c^{\frac1p+1}}{1+p}\frac{\Gamma\left(\frac12-\frac1{2p}\right)\Gamma\left(\frac1{2p}\right)}{2\sqrt\pi}-\frac{1}{1+p}\sum_{k=0}^\infty\binom{-\frac12}{k}\frac{c^{2k+2}}{2k+1-\frac1p}\,\,}\tag{4}$$ We can write out several first terms explicitly: $$I(p,c)\sim\frac{1+\frac{c^2}{2}-\frac{c^4}{8}+\frac{c^6}{16}-...}{1+p}+\frac{c^{\frac1p+1}}{1+p}\frac{\Gamma\left(\frac12-\frac1{2p}\right)\Gamma\left(\frac1{2p}\right)}{2\sqrt\pi}$$ $$-\,\frac{p\,c^2}{(p+1)(p-1)}+\frac12\frac{p\,c^4}{(p+1)(3p-1)}-\frac38\frac{p\,c^6}{(p+1)(5p-1)}+...\tag{5}$$ We found asymptotics for $p>1$, but in fact it is valid for $p>0$, as the initial integral is well-defined for $p>0$. We can extend (4) to $p\in(0;1]$, handling the gamma-function of negative argument and disclosing uncertainties at the "dangerous" points $\displaystyle p=1, \frac13,\frac15,...\frac1{2k+1}...$

For example, at $p\to1$ we have the singular part $$\frac{c^{\frac1p+1}}{1+p}\frac{\Gamma\left(\frac12-\frac1{2p}\right)\Gamma\left(\frac1{2p}\right)}{2\sqrt\pi}\to\frac{c^2}{4}\frac{\Gamma\left(\frac12-\frac1{2p}\right)\left(\frac12-\frac1{2p}\right)}{\left(\frac12-\frac1{2p}\right)}\to\frac{c^2}{2}\frac{p\,\Gamma\left(\frac32-\frac1{2p}\right)}{p-1}\to\frac{c^2}{2(p-1)}$$ and this singularity (at $p\to1$) is canceled by the term $\displaystyle -\,\frac{p\,c^2}{(p+1)(p-1)}$ in (5), so we get only regular terms.

One can easily check, for example, that at $p\to\frac13$ the singularity is canceled by the term $\displaystyle \frac12\frac{p\,c^4}{(p+1)(3p-1)}$, etc.

One more check: at $p\to\infty \,\,I(\infty,c)=c$, what is clear from the initial integral. On the other hand, all the terms in (4) except for $\,\frac{c^{\frac1p+1}}{1+p}\frac{\Gamma\left(\frac12-\frac1{2p}\right)\Gamma\left(\frac1{2p}\right)}{2\sqrt\pi}$ vanish, and $$\frac{c^{\frac1p+1}}{1+p}\frac{\Gamma\left(\frac12-\frac1{2p}\right)\Gamma\left(\frac1{2p}\right)}{2\sqrt\pi}\to \frac{c\,\Gamma\big(\frac12\big)}{\sqrt\pi}\frac1{2p}\Gamma\big(\frac1{2p}\big)\to c$$

Answer 2

Expand the integrand with the generalized binomial theorem and switch the integral and the series with the Fubini-Tonelli theorem:

$$F(c)= \int_{0}^{1} \sum_{j=0}^{\infty} \binom{\frac{1}{2}}{j} x^{2p(\frac{1}{2}-j)}c^{2j} dx = \sum_{j=0}^{\infty} \binom{\frac{1}{2}}{j} c^{2j}\int_{0}^{1}x^{p(1-2j)} dx = \sum_{j=0}^{\infty} \binom{\frac{1}{2}}{j}\frac{1}{p(1-2j)+1} c^{2j}$$

This is true if $|x^{2p}|>|c^{2}|$ which means that $-x^{p}<c<x^{p}$ or $|c|<1 $

Note that the coefficient of $c^{1}$ is zero given that the powers of $c$ in the expansion are even.

Answer 3

If you are familiar with the Gaussian hypergeometric function $$\int \sqrt{x^{2p}+c^2}\,dx=c \,x \,\,\, _2F_1\left(-\frac{1}{2},\frac{1}{2 p};\frac{2p+1}{2 p};-\frac{x^{2 p}}{c^2}\right)$$ Assuming $c>0$ $$\int_0^1 \sqrt{x^{2p}+c^2}\,dx=c \,\,\, _2F_1\left(-\frac{1}{2},\frac{1}{2 p};\frac{2p+1}{2 p};-\frac{1}{c^2}\right)$$ Expanding as series around $c=0$, the first terms would be $$F(c)=\frac{1}{p+1}-\frac{c^2}{2(p-1)}+\frac{c^4}{8 (3 p-1)}-\frac{c^6}{16(5 p-1)}+\frac{5 c^8}{128 (7 p-1)}+O\left(c^{10}\right)+$$ $$\frac{1}{2 \sqrt{\pi }} \Gamma \left(\frac{2p+1}{2 p}\right) \Gamma \left(-\frac{p+1}{2 p}\right)c^{\frac{p+1}{p}}$$ I am sure that you see the patterns as well as the potential problems already mentioned by @Svyatoslav