Taylor expansion of $f(x)=\int_0^1 \frac{1}{\sqrt{t^3+x^3}}dt$ Let $f(x)=\int_0^1 \frac{1}{\sqrt{t^3+x^3}}dt$, for $x>0.$
Show that for all $n\in \mathbb N$, the function $g(x)=f(x)+\ln(x)$ has an expansion of order $n$. Comptue for $n=8$.

I have tried to develop the term $\frac{1}{\sqrt{t^3+x^3}}$ in the neihborhood of $0$ but in vain. Any help is really appreciated.
 A: The integral can be written using a hypergeometric function as
$$f(x) = x^{3/2} {}_{2}^{}{{{F_{1}^{}}}}\! \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{x^{3}}\right)$$
According to Maple, this has the Puiseux series
$$ \frac{4 \pi^{2} 2^{\frac{2}{3}}}{9 \Gamma \! \left(\frac{2}{3}\right)^{3} \sqrt{x}}-2+\frac{x^{3}}{7}-\frac{3 x^{6}}{52}+\frac{5 x^{9}}{152}+O\! \left(x^{12}\right)$$
So it's not $\ln(x)$ that you need to add, but a constant times $x^{-1/2}$.
I think the question was intended to use a cube root rather than square root.
A: Assume for a moment that $0 < x < 1$, and write
\begin{align*}
f(x)
&= \int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}}
- \int_{1}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}}.
\end{align*}
The first integral can be handled by substituting $t = x u^{1/3}$:
\begin{align*}
\int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}}
&= \frac{1}{3\sqrt{x}} \int_{0}^{\infty} \frac{u^{1/3-1}}{(1 + u)^{1/2}} \, \mathrm{d}u \\
&= \frac{1}{3\sqrt{x}} B\left(\frac{1}{3}, \frac{1}{6} \right)
= \frac{\Gamma(4/3)\Gamma(1/6)}{\Gamma(1/2)\sqrt{x}}.
\end{align*}
The second integral can be integrated term-by-term using the binomial series:
\begin{align*}
\int_{1}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}}
&= \int_{1}^{\infty} \frac{\mathrm{d}t}{t^{3/2} \sqrt{1 + (x/t)^3}} \\
&= \int_{1}^{\infty} \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{x^{3n}}{t^{3n+3/2}} \\
&= \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{x^{3n}}{3n+1/2} \\
&= 2 - \frac{x^3}{7} + \frac{3 x^6}{52} - \frac{5 x^9}{152} + \cdots.
\end{align*}
Therefore, we get
$$ f(x) = \frac{\Gamma(4/3)\Gamma(1/6)}{\Gamma(1/2)\sqrt{x}} - \left( \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{x^{3n}}{3n+1/2} \right). $$
This series is the same as what Robert Israel found.
