I am trying to prove results (a) and (b) for definite integrals, and have to carefully justify my steps. My question is NOT how to show these results can be obtained in other ways. The question is how do I justify switching the infinite sum and integral in proving (b). $$(a) \int_0^\infty \frac{t}{e^t -1} dt= \frac{\pi^2}{6} , \\ (b) \int_0^\infty \frac{t}{e^t +1} dt= \frac{\pi^2}{12} $$ For (a) I showed $$\sum_{k=1}^\infty \frac{1}{k^2} = \sum_{k=1}^\infty \int_0^\infty t e^{-kt} dt = \int_0^\infty t\sum_{k=1}^\infty (e^{-t})^k dt = \int_0^\infty \frac{te^{-t}}{1- e^{-t}} dt .$$ The sum and integral can be switched by the Monotone Convergence Theorem since the partial sums have nonnegative terms and, therefore, converge in an increasing fashion. For (b) I can follow similar steps but now I need to justify: $$= \sum_{k=1}^\infty \int_0^\infty (-1)^{k+1} t e^{-kt} dt = \int_0^\infty t \sum_{k=1}^\infty (-1)^{k+1} (e^{-t})^k dt .$$ the convergence is not monotone and the series does not converge uniformly on $(0,\infty).$ Thank you for your help.
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You can use the dominated convergence theorem. For every $n \geqslant 1$ we have
$$0 \leqslant \sum_{k = 1}^n (-1)^{k+1} e^{-kt} \leqslant e^{-t},$$
since the sum is alternating and $e^{-kt}$ decreases with $k$.
Daniel Fischer
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Note that if $\int (\sum_k |f_k|) < \infty,$ then
$$\sum_k \int f_k = \int (\sum_k f_k ).$$
This follows from the dominated convergence theorem. In your problem we have the hypothesis in the first line satisfied (as in part (a)).
zhw.
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