Can I invert this expression with an integral transform?

Question

Let $$ T(t) = 1 - \int_{\tau = 0}^\infty a(\tau) \mathrm{e}^{-t/\tau} d\tau. $$where $t \in (0, \infty)$. I know $ T(t) $ and would like to find $ a(\tau) $. This looks like a Laplace transform but the limits of integration are the wrong way round. Can I find $ a(\tau) $ using an integral transform?

Answer 1

$T(t)=1-\int_0^\infty a(\tau)e^{-\frac{t}{\tau}}~d\tau$

$\int_\infty^0a\left(\dfrac{1}{\tau}\right)e^{-t\tau}~d\left(\dfrac{1}{\tau}\right)=1-T(t)$

$\int_0^\infty\dfrac{1}{\tau^2}a\left(\dfrac{1}{\tau}\right)e^{-t\tau}~d\tau=1-T(t)$

$\mathcal{L}_{\tau\to t}\left\{\dfrac{1}{\tau^2}a\left(\dfrac{1}{\tau}\right)\right\}=1-T(t)$

$\dfrac{1}{\tau^2}a\left(\dfrac{1}{\tau}\right)=\delta(\tau)-\mathcal{L}^{-1}_{t\to\tau}\left\{T(t)\right\}$

$a\left(\dfrac{1}{\tau}\right)=\tau^2\delta(\tau)-\tau^2\mathcal{L}^{-1}_{t\to\tau}\left\{T(t)\right\}$

$a(\tau)=\dfrac{1}{\tau^2}\delta\left(\dfrac{1}{\tau}\right)-\dfrac{1}{\tau^2}\mathcal{L}^{-1}_{t\to\frac{1}{\tau}}\left\{T(t)\right\}$