Dominated convergence theorem

If $f_n\to f$ almost everywhere and $|f_n|\le g$ for one integrable function $g$, then $f$ is integrable and $\int |f_n-f|\to 0$.

For real $a$, the geometric-series expansion of $1/(e^x-1)$ gives an integrable series after multiplication by $\sin (ax)$: each absolute integral is bounded by $|a|{\int }_0^{\infty }x{e}^{-nx}dx=|a|/n^2$. Termwise integration is therefore justified and gives the convergent series ${\sum }_{n\ge 1}a/(a^2+n^2)$.

If $f$ is jointly continuous on a compact rectangle $J\times [a,b]$, then the displayed parameter integral is continuous on $J$. For $t_n\to t$, pointwise continuity gives $f(t_n,x)\to f(t,x)$. The extreme value theorem bounds $|f|$ by a constant $M$ on the rectangle, and $M$ is integrable on the finite interval $[a,b]$. The dominated convergence theorem gives $F(t_n)\to F(t)$.

The same conclusion holds on an unbounded parameter interval if $f$ is jointly continuous there: use a compact rectangle around each fixed parameter value. A global bound over the whole parameter range is unnecessary.

Suppose $u$ and $\log (1+x)u$ are Lebesgue integrable functions on $(0,\infty )$. For integers $n\ge 1$, $x+n\le n(1+x)$ gives the displayed L1 norm bound. It tends to zero because $\log n/n\to 0$, proving L1 convergence of the logarithmically weighted functions to zero. Alternatively, $|\log (n+x)u(x)/n|\le (1+\log (1+x))|u(x)|$ and pointwise convergence permit the dominated convergence theorem.

For $u(x)=\sin x/x^q$ with $1<q<2$, both integrability conditions hold. Near zero, $|\sin x|\le x$ and $\log (1+x)\le x$ give integrable bounds by powers. At infinity, $|\sin x|\le 1$ and $\log (1+x)\le \log 2+\log x$, with ${\int }_1^{\infty }{x}^{-q}\log xdx=(q-1{)}^{-2}$. The L1 norm consequently decays as $O(\log n/n)$ in Big O notation.