Interchanging integration with summation on a specific example

Question

Consider the infinite sum $$S=\sum_{j=1}^\infty (-1)^{j +1}\frac{1}{2j+1} \int_0^1 \frac{1-x^{2j}}{1-x}dx$$ and the associated integral $$L=\int_0^1 \sum_{j=1}^\infty\frac{ 1-x^{2j}}{1-x} (-1)^{j +1}\frac{1}{2j+1} dx.$$ Using in $L$ the power series for the $\arctan$ function, one can show that
$$L= \int_0^1 \frac{\arctan x -(\pi/4) x}{x(1-x)} dx,$$ and of course $S=\sum_{j=1}^\infty (-1)^{j +1} \frac{H_{2j}}{2j+1}$. Combined with some manipulations, the standard softwares yield that $S=L=\frac{\pi}{8} \log 2$. My question: why one can interchange in the formulas for $S$ and $L$ integration with summation? We were unable to find an integrable upper bound for the partial sums for the integrand of $L$. Moreover the series representing the integrand for $L$ definitely does not converge uniformly on $(0,1)$. Due to a lack of monotonicity in $j$ of the summands, the Leibniz rule for alternating series cannot be applied either. So is there any other argument garantying that $\int\sum=\sum\int$?

Answer 1

Fubini's theorem states that, if the absolute value of the integrand/summand doesn't change the convergence properties of the expression, then you could switch symbols. In this case, Fubini's Theorem allows switching the integral and summation symbol.

Answer 2

The series $\sum_{j=1}^\infty\frac{(-1)^{j +1}}{2j+1}\frac{ 1-x^{2j}}{1-x} $ is uniformly convergent on $[0,1-\delta]$ for any $\delta \in (0,1)$ as a consequence of Abel's Test.

Thus,

$$\tag{*}\int_0^1 \sum_{j=1}^\infty\frac{(-1)^{j +1}}{2j+1}\frac{ 1-x^{2j}}{1-x} \, dx \\ =\lim_{\delta \to 0+}\int_0^{1-\delta}\sum_{j=1}^\infty\frac{(-1)^{j +1}}{2j+1}\frac{ 1-x^{2j}}{1-x} \, dx \\= \lim_{\delta \to 0+}\sum_{j=1}^\infty\frac{(-1)^{j +1}}{2j+1}\int_0^{1-\delta}\frac{ 1-x^{2j}}{1-x} \, dx\\ = \lim_{\delta \to 0+}\sum_{j=1}^\infty\frac{(-1)^{j +1}}{2j+1}[\underbrace{(1-\delta) + \frac{(1-\delta)^2}{2} + \cdots +\frac{(1-\delta)^{2j}}{2j}}_{\hat{H}_{2j}(1-\delta)}]$$

It is not difficult to show that the sequence $\frac{\hat{H}_{2j}(1-\delta)}{2j+1}$ is decreasing and convergent to $0$. We also have

$$0 < \frac{\hat{H}_{2j}(1-\delta)}{2j+1}\leqslant \frac{H_{2j}}{2j+1},$$

where $H_{2j}$ is the $2j$-th harmonic number. This implies that the convergence of $\frac{\hat{H}_{2j}(1-\delta)}{2j+1}$ is uniform for $\delta \in [0,1)$. Since the partial sums $\sum_{j=1}^n(-1)^{j+1}$ are uniformly bounded, it follows from the Dirichlet test that the series on the RHS of (*) is uniformly convergent for $\delta \in [0,1)$.

Consequently, we can exchange the limit and the sum in (*) to obtain

$$\int_0^1 \sum_{j=1}^\infty\frac{(-1)^{j +1}}{2j+1}\frac{ 1-x^{2j}}{1-x} \, dx =\sum_{j=1}^\infty\frac{(-1)^{j +1}}{2j+1}\int_0^{1}\frac{ 1-x^{2j}}{1-x} \, dx$$