Construct a Taylor Polynomial of a function containing an integral

Question

I am trying to create a taylor polynomial (degree 5) of

$(1/x)\int_{0}^{x}e^{-v^{2}} dv$

Now I've created taylor polynomials before, but I've never seen a problem involving two variables.

Answer 1

Let $$f(x) = \frac{1}{x}\int_{0}^{x}e^{-t^2}dt$$ then $$f'(x) = \frac{-1}{x^2}\int_{0}^{x}e^{-t^2}dt+\frac{e^{-x^2}}{x} = \frac{-f(x)}{x}+\frac{e^{-x^2}}{x}.$$ This will give you a recurrence relation to generate successive derivatives with which you could construct your polynomial.

Or something even better, you could change variables and write $$f(x) = \frac{1}{x}\int_{0}^{x}e^{-t^2}dt$$ as $(t= xu)$ $$f(x)= \int_{0}^{1}e^{-x^2u^2}du = \int_{0}^{1}\sum_{k=0}^{\infty} (-1)^{k}\frac{(xu)^{2k}}{k!}du\\ = \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k+1)k!}.$$

Answer 2

$$e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$$ $$e^{-t^2}=\sum_{n=0}^{\infty}\frac{(-1)^{n}(t)^{2n}}{n!}$$ $$\int_{0}^{x}e^{-t^2}dt=\sum_{n=0}^{\infty}\frac{(-1)^{n}(x)^{2n+1}}{(2n+1)n!}$$ so $$\frac{1}{x}\int_{0}^{x}e^{-t^2}dt=\sum_{n=0}^{\infty}\frac{(-1)^{n}(x)^{2n}}{(2n+1)n!}$$