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