Use a power series to approximate the definite integral, I, to six decimal places.
I tried to definite integral, but the answer is wrong. Where did I make a mistake?
$$\int_{0}^{0.2} \ln(1+x^4)\ dx = \int_{0}^{0.2} \sum_{n=1}^{\infty} \frac {(-1)^{n+1}\cdot x^{4n}}{n(4n+1)}\ dx
= \left[\frac {x^5}{5} – \frac {x^9}{18} +\frac {x^{13}}{39} – \frac {x^{17}}{68} … \right]\bigg|_{0}^{0.2} $$
I used the first 3 terms, the error is at most $$\frac {0.2^{17}}{68} $$
So, $$I \approx 0.000060 $$
The answer is wrong! How do I figure it out? Thank you.
Best Answer
Hint. One may use the Taylor series expansion $$ \log (1+u)=u-\frac{u^2}2+\frac{u^3}3-\frac{u^4}4+\frac{u^5}5+\cdots, \quad |u|<1, $$ giving, for $0<a<1$, $$ \int_0^a\log (1+x^4)\:dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^a\frac{x^{4n}}n\:dx=\sum_{n=1}^\infty(-1)^{n-1}\frac{a^{4n+1}}{n(4n+1)}. $$ Then, for $0<a<1$, using the classic inequality (for alternating series) $$ \left|\sum_{n=n_0}^\infty(-1)^{n-1}\frac{a^{4n+1}}{n(4n+1)}\right|\leq \frac{a^{4n_0+1}}{n_0(4n_0+1)}, \quad n_0\geq1, $$ you get $$ \left|\int_0^a\log (1+x^4)\:dx-\sum_{n=1}^{n_0}(-1)^{n-1}\frac{a^{4n}}{n(4n+1)}\right|\leq\frac{a^{4n_0+1}}{n_0(4n_0+1)}. $$ By applying it to $a=0.2$ and $n_0=2$, we obtain $$ \left|\int_0^{0.2}\log (1+x^4)\:dx-\sum_{n=1}^2(-1)^{n-1}\frac{(0.2)^{4n}}{n(4n+1)}\right|\leq\frac{(0.2)^9}{2\times9}\approx 0.00000003. $$