Parseval formula for the Mellin transform

Remarks

The integrability hypotheses say that $f$ is Mellin convergent at $\sigma$ and $g$ at $1 - \sigma$, so the transforms on the right are defined pointwise. Put $h_f(u) = e^{\sigma u} f(e^{u})$ and $h_g(u) = e^{(1 - \sigma) u} g(e^{u})$. The substitution $x = e^{u}$ turns the hypotheses into $h_f, h_g \in L^1(\R) \cap L^2(\R)$ and the left sides into $\int \abs{h_f}^2 \, du$ and $\int h_f h_g \, du$. It also turns $\mathcal{M}f(\sigma + i\tau)$ and $\mathcal{M}g(1 - \sigma - i\tau)$ into Fourier transforms of $h_f$ and $h_g$ at $\mp \tau / 2\pi$. The Plancherel theorem for the Fourier transform on $\R$ then gives both formulas. Example: $f(x) = e^{-x}$ and $\sigma = 1/2$ give $\int_0^\infty e^{-2x} \, dx = \frac{1}{2} = \frac{1}{2\pi} \int_{\R} \abs{\Gamma(\tfrac12 + i\tau)}^2 \, d\tau$; since $\abs{\Gamma(\tfrac12 + i\tau)}^2 = \pi / \cosh(\pi \tau)$, the right side is $\frac12 \int_{\R} \frac{d\tau}{\cosh \pi \tau} = \frac12$. By density, part 1 extends $\mathcal{M}$ to a unitary map, up to the factor $2\pi$, from $L^2((0, \infty), x^{2\sigma - 1} dx)$ onto $L^2(\R)$.

Going deeper

For a student

The formula is used to evaluate integrals of products: one knows $\mathcal{M}f$ and $\mathcal{M}g$, and computes $\int_0^\infty f g \, dx$ by a single integral on a vertical line, often by residues. Part 2 can also be read as the convolution rule for the Mellin convolution of $f$ and $x \mapsto g(1/x)/x$, combined with Mellin inversion at the point $x = 1$. On the critical line $\sigma = 1/2$ the weight disappears, and part 1 says that $f \mapsto \mathcal{M}f(\tfrac12 + i\tau) / \sqrt{2\pi}$ preserves the norm of $L^2(0, \infty)$.