Find an asymptotic expansion at order $6$ of $f(x) = \int_x^{x^2} \frac{\mathrm{d}t}{\sqrt{1+t^4}}$
I don't know how to proceed. I think I need to do a change of variable yet I don't know which one. I tried $u = t/x$ yet it doens't seem to work...
Thank you !
On
You want $f(x) = \int_x^{x^2} \frac{\mathrm{d}t}{\sqrt{1+t^4}} $ to order 6.
Being as general as possible, let $f(x) = \int_{x^a}^{x^b} (1+t^u)^v dt $.
Then, using the generalized binomial theorem,
$\begin{array}\\ f(x) &= \int_{x^a}^{x^b} (1+t^u)^v dt\\ &= \int_{x^a}^{x^b} \sum_{n=0}^{\infty} \binom{v}{n}t^{un}dt\\ &= \sum_{n=0}^{\infty} \binom{v}{n}\int_{x^a}^{x^b} t^{un}dt\\ &= \sum_{n=0}^{\infty} \binom{v}{n}\dfrac{t^{un+1}}{un+1}|_{x^a}^{x^b} \\ &= \sum_{n=0}^{\infty} \binom{v}{n}\dfrac{(x^b)^{un+1}-(x^a)^{un+1}}{un+1} \\ &= \sum_{n=0}^{\infty} \binom{v}{n}\dfrac{x^{b(un+1)}-x^{a(un+1)}}{un+1} \\ \end{array} $
Then take as many terms as you want.
I'm assuming you are looking for an expansion as $x\rightarrow 0$.
For $u\rightarrow 0$, $$(1+u)^\alpha = 1 + \alpha u + \mathcal O(u^2)$$ Therefore $$\frac 1 {\sqrt{1+t^4}}=1 -\frac 1 2 t^4 +\mathcal O(t^{8})$$ and $$\begin{split} f(x) &= \int_x^{x^2} \frac{\mathrm{d}t}{\sqrt{1+t^4}}\\ &=\int_x^{x^2}\left(1-\frac 1 2 t^4+\mathcal O\left(t^8\right)\right)dt\\ &= (x^2-x)-\frac 1 2\left(\frac{x^{10}}{5}-\frac{x^5}5\right)+\mathcal O\left(x^9\right)\\ &=-x +x^2+\frac{x^5}{10}+\mathcal O\left(x^9\right) \end{split}$$