Taylor Expansion to Solve Definite Integral within Specific Accuracy

Question

I have a problem that asks Evaluate to five decimal places using the taylor series for the definite integral. $$\int_{1}^{2} \frac{e^x}{x} dx$$

I am not sure how to set this up. I know the Taylor expansion is

$$\frac{e^x}{x}=\frac{1}{x}+1+\frac{x}{2!}+\frac{x^2}{3!}+....+\frac{x^{n-1}}{(n)!}+\frac{x^{n}e^{\xi}}{(n+1)!}$$

So now we integrate the problem and we have

$$\left.ln(x)+x+\frac{x^2}{4}+\frac{x^3}{(3!)(3)}+...+\frac{x^{n}}{(n!)(n)}\right|_1^2+\int_{1}^{2}\frac{x^{n}e^{\xi}}{(n+1)!} dx$$

Now, the below is the Taylor remainder of this series, and it needs to be upper bounded by the term so we have: $$\int_{1}^{2}\frac{x^{n}e^{\xi}}{(n+1)!}dx < 5.0x10^{-6}$$.

I already have solved one of these that involved a=0 and b=1 but I do not know how to do this with non zero and 1. This means the error term is

$$\left.\frac{x^{n+1}e^{\xi}}{(n+1)!(n+1)}\right|_1^2$$

When I was dealing with 0 and 1 the error was bounded by $$\left.\frac{x^{n}e^{\xi}}{(n+1)!(n+1)}\right|_0^1$$ which gave us

$$\frac{e^{}}{(n+1)!(n+1)}<5.0x10^{-6} $$ which meant when n=7 the inequality was solved so we took the the first 8 terms of the Taylor series to solve within that accuracy. However, what do I do when the terms is below. $$\left.\frac{x^{n+1}e^{\xi}}{(n+1)!(n+1)}\right|_1^2$$

IF someone could point me what is the actual inequality I am trying to solve I would greatly appreciate it. From my book, since I am using the taylor series the $\xi$ has to be between 0 which what my series starts as and what x is. the problem I already solved it was between 0 and 1 and the integral was between those as well. However, now, with this, it would be between zero and 2 then and bounded by two for the larger if I am making the same conjecture. Again, at first I thought the below would be the inequality that I would solve but I am not sure at this point.

$$\frac{2^{n+1}e^{2}}{(n+1)!(n+1)}-\frac{e}{(n+1)!(n+1)}<5.0x10^{-6}$$ Find when n is true? this would be n=11

Answer 1

Note that your error term is $$\int_{1}^{2}\frac{x^{n}e^{\xi}}{(n+1)!} $$ and you want to find an upper bound for it.

The upper bound is found by multiplying the length of your interval which is $2-1 =1$ by the maximum value of your integrand.

Your maximum value of the integrand is $$\frac{2^{n}e^{2}}{(n+1)!} $$ which is also the upper upper bound for your error.

You can continue from here.

Answer 2

Since most of the mass of the integral is concentrated around $x=2$, a faster-converging (double-)series representation for the given integral can be derived in the following way:

$$ I = \frac{e^2}{2}\int_{0}^{1}\frac{e^{-x}}{1-\frac{x}{2}}\,dx=\frac{e^2}{2}\sum_{n\geq 0}\frac{1}{2^n}\int_{0}^{1}x^n e^{-x}\,dx=\frac{e^2}{2}\sum_{n\geq 0}\frac{1}{2^n}\sum_{m\geq 0}\frac{(-1)^m}{(n+m+1)m!}.\tag{1} $$ An accuracy within $10^{-6}$ can be achieved the truncating the RHS of $(1)$ up to $n=20$ and $m=10$, leading to $$ I\approx \frac{26724910640645913406379 e^2}{64551926790503792640000} = \color{green}{3.059116}494\ldots\tag{2} $$ The actual value of $I$ is $3.0591165396459534\ldots$

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A slicker way is to exploit Beuker-like integrals. $I=\text{Ei}(2)-\text{Ei}(1)$, by integration by parts, is related to $$ \int_{1}^{2}\frac{e^x}{x}(x-1)^8 (2-x)^8\,dx $$ whose value is extremely close to zero. It follows that $$ I \approx -9275892499 e+\frac{54598562389 e^2}{16} = \color{green}{3.059116}486\ldots\tag{3}$$