After this question: $$\int\ln(1+ae^{bx})\\=\int\left(ae^{bx}-\frac{a^2e^{2bx}}2+\frac{a^3e^{3bx}}3+...\right)\\=-\frac1b\left(\frac{-ae^{bx}}1+\frac{a^2e^{2bx}}{4}-\frac{a^3e^{3bx}}{9}...\right)\\=-\frac1b{\rm Li}_2(-ae^{bx})$$ If this is wrong, Can you suggest me a how to improve the proof/a better proof?
Asked
Active
Viewed 81 times
2
-
No, you can't integrate just a piece of the function ignoring the dots, unless you are only going for an approximate $O(x^n)$-type result (in which case you don't need special functions). – Mario Carneiro Feb 06 '15 at 14:56
-
According to a CAS, your result seems to be the good one ! Cheers – Claude Leibovici Feb 06 '15 at 14:56
-
@Mario Carneiro: I think that, rather than terms being dropped, there is a missing "$+\cdots$" in the third expression where the denominators are squares and the sign alternates. There may also be a missing "^" in the fourth expression. – Henry Feb 06 '15 at 15:08
-
@Henry Even if that were the case, I don't see how one goes from an equation with omitted terms to one without them and not make mistakes in the proof unless the omitted terms are explicitly described by a summation or "general term" among the dots. Even if the answer is correct, this is not a proof by any stretch. – Mario Carneiro Feb 06 '15 at 15:10
-
@Henry yes there was those typos – RE60K Feb 06 '15 at 15:18
1 Answers
3
It might be clearer if you wrote the series for $\log(1+x)$ as an infinite series: $$ \begin{align} \int\log(1+ae^{bx})\,\mathrm{d}x &=\int\sum_{k=1}^\infty(-1)^{n-1}\frac{a^n}ne^{nbx}\,\mathrm{d}x\\ &=\frac1b\sum_{k=1}^\infty(-1)^{n-1}\frac{a^n}{n^2}e^{nbx}+C\\ &=-\frac1b\,\mathrm{Li}_2\!\left(-ae^{bx}\right)+C \end{align} $$ Otherwise, your answer looks correct.
robjohn
- 345,667