# Taylor series leads to two different functions - why?

Suppose, I want to find a function such that its Taylor series expansion is $$f(x) = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)a^n}$$

I could start with $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$

Integrate it, substitute $$x\rightarrow \frac{x}{a}$$, multiply by $$a$$ and get

$$F(x) = -\ln|x-1| = \sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$$

$$a F\left(\frac{x}{a}\right) = -a \ln\left|\frac{x}{a}-1\right| = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)a^n}$$

On the other hand, I could start with subtituting $$x \rightarrow \frac{x}{a}$$ before integration to get

$$\frac{a}{a-x} = \sum_{n=0}^{\infty}\frac{x^n}{a^n}$$

and then integrate it to get $$-a\ln|x-a| = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)a^n}$$

As you can see, arguments of $$\ln$$ are not equal. Where did it go wrong?

• it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 \ \ f'(x)=g'(x) = 4x$ – Jneven May 16 at 8:51
• It's good that you ask this question. But that's why it always pays to be precise. If you had attempted to define exactly what $F$ was, you would have realized that either you define it as some anti-derivative, or you define it as some specific definite integral, and in both cases you will know that you have to handle the constants that arise correctly. – user21820 May 16 at 9:37

## 1 Answer

When you integrate, you should include a constant of integration. What you see here is that when integrating the functions, you get different constants of integration. This is why your answers differ by only a constant, namely $$a\ln a$$ (you can see this by use of $$\log$$ rules).

If you take care with the limits or boundary conditions in the integration step, then the answers will agree exactly.