What is the sum of 1(1 - 1/3 ) - 1/2(1 - 1/3 + 1/5 - 1/7)+1/3(1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11) - 1/4(...)...?

Question

The series converges (conditionally) since $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}+\ldots $$ converges to $\ln{2}$, and $$ 1 - \frac{1}{3},\quad 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7},\quad 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11}... $$is a monotonically increasing sequence with a limit of $\frac{\pi}{4}$. However I have made no progress at all in finding an expression for its sum. Any suggestions or references would be appreciated.

Answer 1

As Winthers has pointed out, the required sum $S$ is given by the following integral $$S = \int^1_0 \frac{\ln 2 - \ln (1 + x^4)}{1 + x^2} \, dx = \frac{\pi}{4} \cdot \ln (2) - \int^1_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx = \frac{\pi}{4} \cdot \ln (2) - J.$$

I will now find the integral $J$ using the most elementary means possible.

We begin by considering the integral $\displaystyle{\int^\infty_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx}$. Writing this as $$\int^\infty_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx = \int^1_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx + \int^\infty_1 \frac{\ln (1 + x^4)}{1 + x^2} \, dx.$$ If in the right most integral we set $x \mapsto 1/x$ one obtains \begin{align*} \int^\infty_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx &= \int^1_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx + \int^1_0 \frac{ \ln \left (1 + \frac{1}{x^4} \right )}{1 + \frac{1}{x^2}} \frac{dx}{x^2}\\ &= 2 \int^1_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx - 4 \int^1_0 \frac{\ln x}{1 + x^2} \, dx. \end{align*}

Now for the right most integral appearing above, integrating by parts yields $$\int^1_0 \frac{\ln x}{1 + x^2} \, dx = - \int^1_0 \frac{\tan^{-1} x}{x} \, dx = - \mathbf{G},$$ where $\mathbf{G}$ is Catalan's constant and comes from one of the well-known integral representations for this constant. Thus, after rearranging, we have $$J = \int^1_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx = \frac{1}{2} \int^\infty_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx - 2 \mathbf{G}.$$ Now it remains to determine the integral $\displaystyle{\int^\infty_0 \frac{\ln (1 + x^4)}{1 + x^2} \, dx}$.

Let $$I(a) = \int^\infty_0 \frac{\ln (1 + ax^4)}{1 + x^2} \, dx, \quad a \geqslant 0.$$ Our required integral is recovered on setting $a = 1$. Differentiating under the integral sign with respect to the parameter $a$, we have $$I'(a) = \int^\infty_0 \frac{x^4}{(1 + ax^4)(1 + x^2)} \, dx,$$ which, under a partial fraction decomposition becomes $$I'(a) = \frac{1}{a + 1} \int^\infty_0 \frac{dx}{1 + x^2} + \frac{1}{a + 1} \int^\infty_0 \frac{x^2}{1 + ax^4} \, dx - \frac{1}{a + 1} \int^\infty_0 \frac{dx}{1 + a x^4}.$$ Finding these three integrals. The first is trivial. It is $$I_1 = \frac{\pi}{2}.$$ The second and third can be found by converting them into the form of beta functions.

For the second integral, on setting $t = a x^4$ it may be rewritten as \begin{align*} I_2 &= \frac{1}{4 a^{3/4}} \int^\infty_0 \frac{t^{-1/4}}{1 + t} \, dt\\ &= \frac{1}{4 a^{3/4}} \int^\infty_0 \frac{t^{3/4 - 1}}{(1 + t)^{3/4 + 1/4}} \, dt\\ &= \frac{1}{4 a^{3/4}} \text{B} \left (\frac{3}{4}, \frac{1}{4} \right ). \end{align*} And since $$\text{B} \left (\frac{3}{4}, \frac{1}{4} \right ) = \Gamma \left (\frac{3}{4} \right ) \Gamma \left (\frac{1}{4} \right ) = \Gamma \left (\frac{1}{4} \right ) \Gamma \left (1 - \frac{1}{4} \right ) = \frac{\pi}{\sin \left (\frac{\pi}{4} \right )} = \pi \sqrt{2},$$ where use has been made of Euler's reflection formula, we arrive at $$I_2 = \frac{\pi \sqrt{2}}{4 a^{3/4}}.$$ The third integral can be found in an exactly analogous manner to how the second was found. The result is $$I_3 = \frac{\pi \sqrt{\pi}}{4 a^{1/4}}.$$ So $$I'(a) = \frac{\pi}{2 (a + 1)} + \frac{\pi \sqrt{2}}{4 a^{3/4} (a + 1)} - \frac{\pi \sqrt{2}}{4 a^{1/4} (a + 1)}.$$

We require $I(1)$. As $I(0) = 0$, on integrating up with respect to $a$ from zero to one we have $$I(1) = \frac{\pi}{2} \int^1_0 \frac{da}{a + 1} + \frac{\pi \sqrt{2}}{4} \int^1_0 \frac{da}{a^{3/4} (a + 1)} - \frac{\pi \sqrt{2}}{4} \int^1_0 \frac{da}{a^{1/4} (a + 1)}.$$ Each of these three integrals can be readily found. The first is $$I_\alpha = \int^1_0 \frac{da}{a + 1} = \ln (2).$$ For the second and the third, after a substitution of $a = x^4$ is made we have (see here for its evaluation) $$I_\beta = 4 \int^1_0 \frac{dx}{1 + x^4} = \frac{\pi + \ln (3 + 2 \sqrt{2})}{\sqrt{2}},$$ and (see here for its evaluation) $$I_\gamma = 4 \int^1_0 \frac{1 + x^2}{1 + x^4} \, dx = \frac{\pi - \ln (3 + 2\sqrt{2})}{\sqrt{2}}.$$

So finally we have \begin{align*} \int^\infty_0 \frac{\ln (1 + x^4}{1 + x^2} \, dx &= \frac{\pi}{2} \cdot \ln (2) + \frac{\pi \sqrt{2}}{4} \cdot \frac{\pi + \ln (3 + 2 \sqrt{2})}{\sqrt{2}} - \frac{\pi \sqrt{2}}{4} \cdot \frac{\pi - \ln (3 + 2 \sqrt{2})}{\sqrt{2}}\\ &= \frac{\pi}{2} \cdot \ln (6 + 4\sqrt{2}). \end{align*} Thus $$J = \frac{\pi}{4} \cdot \ln (6 + 4 \sqrt{2}) - 2 \mathbf{G}.$$

So our sum, in closed formed, becomes $$S = \frac{\pi}{4} \cdot \ln (2) - \frac{\pi}{4} \cdot \ln (6 + 4 \sqrt{2}) + 2\mathbf{G} = 2 \mathbf{G} - \frac{\pi}{4} \cdot \ln (3 + 2 \sqrt{2}).$$

Answer 2

\begin{eqnarray*} G=\sum_{i=1}^{\infty} \frac{(-1)^i}{i} \sum_{j=0}^{2i} \frac{(-1)^j}{2j+1} \end{eqnarray*} \begin{eqnarray*} \int_{0}^{1} y^{i} dy =\frac{1}{i} \\ \int_{0}^{1} x^{2j+1} dx = \frac{(-1)^j}{2j+1} \end{eqnarray*} Substitute these & perform the geometric sums, do the y integrals ... we get \begin{eqnarray*} G=\frac{\pi}{4} \ln2- \int_{0}^{1} \frac{\ln(1+x^4)}{(1+x^2)} dx \end{eqnarray*} The value of the integral is roughly $0.09690 \cdots$ & $G=0.4475 \cdots$ (In agreement with Winther). I am not sure how to do this integral at the moment, but given that $1+x^4=(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)$ it certainly give a clue where the logarithmic term in Winthers answer comes from.