a) Find the Maclaurin expansion of the following function: $$f(x)=\int\limits_0^x \frac{1-e^{-t^3}}{t^2} \mathrm{d}t$$ end b) evaluate the $ \displaystyle \lim_{x \to 0^{+}} f^{(29)}\, (x) $
The above problem is from a past exam at calculus II (integral calculus) at my university.
How is it possible to express $ f(x) $as a Macluarin series since $ \displaystyle \frac{1-e^{-t^3}}{t^2} $ does not exists for $ t_0=0 $?
so i would like to see some hints or\and solutions. Thank you in advance.
$$ 1-e^{-t}=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n!}t^n\tag{1} $$ hence by replacing $t$ with $t^3$ and by dividing both terms by $t^2$ we get: $$ \frac{1-e^{-t^3}}{t^2}=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n!}t^{3n-2}\tag{2} $$ then, integrating termwise: $$ \int_{0}^{x}\frac{1-e^{-t^3}}{t^2}\,dt = \sum_{n\geq 1}\frac{(-1)^{n+1}}{(3n-1)n!}x^{3n-1}\tag{3}$$ we get the wanted Taylor series.