= Solution
Use $q=e^{2\pi i\tau}$, so $g(iy)=\sum_{n\geq1}b_ne^{-2\pi ny}$. Put $u=\operatorname{Re}s$. For $u>\max\{0,\sigma+1\}$,
$$
\sum_{n\geq1}|b_n|\int_0^\infty e^{-2\pi ny}y^{u-1}\,dy=\Gamma(u)(2\pi)^{-u}\sum_{n\geq1}|b_n|n^{-u}<\infty.
$$
The <Gamma integral> and <absolute convergence> therefore justify exchanging summation and integration. For every positive integer $n$, substitution $t=2\pi ny$ gives
$$
\int_0^\infty e^{-2\pi ny}y^{s-1}\,dy=(2\pi n)^{-s}\Gamma(s).
$$
Summing these identities proves the <Mellin transform> formula for the <Dirichlet series>:
$$
\boxed{\int_0^\infty g(iy)y^s\frac{dy}{y}=(2\pi)^{-s}\Gamma(s)L(g,s).}
$$
The power of the positive real variable $y$ uses the real logarithm; the initial half-plane stated above is a sufficient one even when $\sigma$ is negative.
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