How about: $ -\int_0^1 (-x)^{i+j-1}\,dx = \frac{(-1)^{i+j}}{i+j} $ then for each $x \in (0,1)$ we have $ \sum_{i=1}^\infty\sum_{j=1}^\infty -(-x)^{i+j-1} = \frac{x}{(x+1)^2} $ and integrate $ \int_0^1\frac{x}{(x+1)^2}\,dx = \log 2 - \frac{1}{2} \approx 0.193147 $
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Explanation for summation inside integral ... Two uses of this nice "monotone alternating" convergence theorem: Suppose $f_1(x) \ge 0\;$ is integrable on $E$ and $f_n(x) \searrow 0$ for almost every $x \in E$. Then $ \sum_{n=1}^\infty (-1)^n \int_E f_n(x)\,dx = \int_E \left(\sum_{n=1}^\infty (-1)^n f_n(x)\right)\,dx $ PROOF: Group the terms in pairs.
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More details now ... $ \int_0^1 -(-x)^{i+j-1}\,dx = \frac{(-1)^{i+j}}{i+j} $ For fixed $i$, the integrand decreases pointwise a.e. to zero in absolute value, and alternates sign. Therefore $ \int_0^1 \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx = \sum_{j=1}^\infty\int_0^1 -(-x)^{i+j-1}\,dx = \sum_{j=1}^\infty \frac{(-1)^{i+j}}{i+j} $ Now this integrand is $ \sum_{j=1}^\infty-(-x)^{i+j-1} = \frac{-(-x)^i}{x+1} $ As $i$ varies, this decreases a.e. to zero in absolute value, and alternates sign, so $ \int_0^1 \sum_{i=1}^\infty \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx= \sum_{i=1}^\infty \int_0^1 \sum_{j=1}^\infty-(-x)^{i+j-1}\,dx= \sum_{i=1}^\infty \sum_{j=1}^\infty \frac{(-1)^{i+j}}{i+j} $