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?

  • 5
    $\begingroup$ it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 \ \ f'(x)=g'(x) = 4x$ $\endgroup$
    – Jneven
    May 16, 2019 at 8:51
  • 6
    $\begingroup$ 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. $\endgroup$
    – user21820
    May 16, 2019 at 9:37

1 Answer 1


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.


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