<s>[INST] Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x\log(1+x^3)=x\int(1-x^3+x^6-x^9+-...)\\\\
&=x\left[x-\frac{x^4}{4}+\frac{x^7}{7}-\frac{x^{10}}{10}+-...+ C\right]
\end{align}$$
$$
f(0)=0; C=0
$$
$$
f(x)=x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...
$$
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)dx&=\int_0^{1/2}(x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...)\,dx\\\\
&=\left.\left[\frac{x^3}{3}-\frac{x^6}{6*4}+\frac{x^9}{7*9}-\frac{x^{12}}{10*12}+-...\right]\right|_0^{1/2}\\\\
&=\frac{1}{2^3(3)}-\frac{1}{2^6(6)(4)}+\frac{1}{2^9(7)(9)}-\frac{1}{2^{12}(10)(12)}+-...
\end{align}$$
Is my method correct?
 [/INST] I would do it this way:
Using
$\ln(1+z)
=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}x^k}{k}
$,
which converges for your range,
$\begin{array}\\
\int_0^{1/2}x\ln(1+x^3)dx
&=\int_0^{1/2}x\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{3k}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int_0^{1/2}x^{3k+1}dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\dfrac{x^{3k+2}}{3k+2}\big|_0^{1/2}\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k(3k+2)2^{3k+2}}\\
&=\dfrac1{5\cdot 2^5}-\dfrac1{2\cdot8\cdot 2^8}+...\\
\end{array}
$
More generally,
$\begin{array}\\
\int x^a\ln(1+x^b)dx
&=\int x^a\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{bk}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int x^{a+bk}dx\\
&=\sum_{k=1}^{\infty} \dfrac{-1)^{k-1}x^{a+1+bk}}{k(a+1+bk)}\\
\end{array}
$
</s>