For $t>0$, the product from part (b) has $q=e^{-2\pi t}\in(0,1)$, so
$$
\Delta(it)>0.
$$
The weight-twelve transformation law gives
$$
\Delta(i/t)=t^{12}\Delta(it).
$$
Together with exponential decay as $t\to\infty$, this implies rapid decay at both endpoints for the <Mellin transform>
$$
I(s)=\int_0^\infty\Delta(it)t^{s-1}\,dt.
$$
The integral therefore converges for every real $s$, and its integrand is strictly positive.

In the half-plane where the Dirichlet series may be integrated term by term,
$$
I(s)=(2\pi)^{-s}\Gamma(s)L(\Delta,s).
$$
Analytic continuation preserves this identity. For real $s>0$, both $(2\pi)^s$ and the <Gamma function> $\Gamma(s)$ are positive, so
$$
L(\Delta,s)=\frac{(2\pi)^s}{\Gamma(s)}I(s)>0.
$$
This is the positivity statement in the <Mellin transform of the modular discriminant>, and in particular $L(\Delta,s)\ne0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 137 / 1 / c / Solution

For every real $s$, the rapidly convergent integral
$$
\int_0^\infty\Delta(it)t^{s-1}\,dt
$$
is the analytic continuation of $(2\pi)^{-s}\Gamma(s)L(\Delta,s)$. The product for $\Delta$ makes the integrand positive, so $L(\Delta,s)>0$ for real $s>0$.


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 Incoming links: Mellin transform of the modular discriminant