I was studying the Linear Algebra perspective about the Laplace Transform. We know that the Laplace Transform is given by:
$$ F(s) = \int_{0}^{\infty}f(t)e^{-st}dt $$
Where $e^{-st}$ is the integral transform kernel. In my point of view, it seems like a change of basis, we changed the coordinates of $f(t)$ to $F(s)$.
But when we apply the Fourier-Mellin Integral, we have:
$$ f(t) = \frac{1}{2\pi i} \oint_{\gamma - i\infty}^{\gamma + i\infty} F(s) e^{st}ds $$ In the $\oint$ the limits are a line in the $\mathbb{C}$ plane around a fix point $\gamma$ in the $\mathbb{R}$ axis with the complex coordinates belonging the interval $(-\infty, \infty)$. So, rewriting the Fourier-Mellin Integral:
$$ f(t) = \int_{\gamma - i\infty}^{\gamma + i\infty} F(s) \left (\frac{e^{st}}{2\pi i} \right) ds $$
So $f(t)$ is a linear combination of functions $B_s(t) = \frac{e^{st}}{2\pi i}$.
Know, I want to prove that $\frac{e^{st}}{2\pi i}$. Applying the definition of orthogonal function: $$ <f(t), g(t)> = \int^{\infty}_{0} f^{*}(t)g(t)dt =0 $$
Where $*$ denotes the complex conjugate.
But the function that I want to proof that is orthogonal is $\frac{e^{st}}{2\pi i}$, so:
$$ <B_s(t), B_{s'}(t)>= \int^{\infty}_{0} \left ( \frac{e^{st}}{2\pi i}\right)^{*} \left ( \frac{e^{s't}}{2\pi i}\right) \\ \implies<B_{s}(t), B_{s'}(t)>=\frac{1}{4 \pi ^{2}} \int_{0}^{\infty} e^{(s^{*}+s)t}dt $$
But in this case: $s = \gamma+iT$. So, replacing $s$ for $\gamma + iT$, we got:
$$ <B_{s}(t), B_{s'}(t)>=\int^{\infty}_{0} e^{(2\gamma + i(T' -T))t}dt $$
At this point, I've got some troubles, because $\gamma$ is the real part of $s$ and the integral needs to converge to the Dirac Delta function:
$$ \delta(x-\alpha)=\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{ip(x-\alpha)}dp $$
To got this result in $<B_{s}(t), B_{s'}(t)>$ I made some changes in the integral:
$$ \gamma \rightarrow 0 $$
And replace the limits as $\infty$ and $-\infty$,
$$ <B_{s}(t), B_{s'}(t)> = \frac{1}{4\pi^{2}} \int_{\infty}^{\infty} e^{i(T'-T)t} dt $$
So the result gives me:
$$ <B_s(t), B_{s'}(t)>=\frac{2\pi}{4\pi^{2}}\delta (T'-T) =\frac{\delta (T'-T)}{\pi} $$
This result, for me, implies that $e^{it}$ is an orthogonal basis once that $\delta$ is $\infty$ when $T'=T$ and $0 $ when $T' \neq T$.
The problem is that I proved $\frac{e^{iT}}{2\pi i}$ is an orthogonal basis instead of $\frac{e^{st}}{2\pi i}$.
How can I prove that $\frac{e^{st}}{2\pi i}$ is an orthogonal basis?