Find the value of this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}….$$
Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\bigg]dx$$
$$S = \int^{1}_{0}(1-x^6)\bigg[1+x^{8}+x^{16}+\cdots \cdots \bigg]dx$$
So $$S = \int^{1}_{0}\frac{1-x^6}{1-x^{8}}dx = \int^{1}_{0}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
Best Answer
HINT: Notice that $$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=\frac{1}{4}\frac{1}{x^2+x\sqrt{2}+1}+\frac{1}{4}\frac{1}{x^2-x\sqrt{2}+1}+\frac{1}{2}\frac{1}{x^2+1}$$ Now we just have to evaluate the integrals $$I_1=\frac{1}{4}\int_0^1 \frac{dx}{x^2+x\sqrt{2}+1}$$ $$I_2=\frac{1}{4}\int_0^1 \frac{dx}{x^2-x\sqrt{2}+1}$$ $$I_3=\frac{1}{2}\int_0^1 \frac{dx}{x^2+1}$$ and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?