Decomposing an intensity spectrum as a superposition of blackbody spectra

Question

My question goes as follows. The same way any integrable function $f(x)$ can be somewhat expressed as a superposition of plane waves as $\int_{-\infty}^{+\infty} F(\lambda)e^{2\pi i x\lambda}d\lambda$, where we know that $F(\lambda)$ can be obtained from $f$ as $\int_{-\infty}^{+\infty} f(x) e^{-2\pi i x\lambda} dx$, if we have the intensity spectrum of any object (say, an astronomical one) as a function of wavelength (or frecuency if you prefer), $I(\lambda)$, can we find a generalised function of the temperature, $A(T)$, such that $I(\lambda) \cong \int_{0}^{+\infty} A(T) B(\lambda; T) dT$, where $B(\lambda; T) = \frac{2hc}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda K_B T}}-1}$ is the spectrum of a blackbody at temperature T?

Physically speaking, we would be somehow looking at the object as a bunch of blackbodies at diferent temperatures somehow coexisting.

We know $I(\lambda)$ is an integrable function going from $(0,\infty)$ to $(0, \infty)$ (and so is $B(\lambda;T)\forall T$, of course), and $I(\lambda\to0)=I(\lambda\to\infty)=0$ (the latter one also being a consequence of $I$ being in $L^1$). An example would be $I(\lambda)=B(\lambda;T_0)$, so $A(T)=\delta(T-T_0)$.

If we can find such an $A$, how do we express it in terms of $I$? If we generally can't, are there some aditional properties $I$ has to have so that $A$ can be obtained?

Answer 1

Ignoring all constants, let

$$\tag{1} B(\lambda,T)=\lambda^{-5}(e^{1/ \lambda T}-1)^{-1} $$

We have the integral equation

$$\tag{2} I(\lambda)=\int\limits_0^\infty dT \ A(T) B(\lambda,T) $$

Where $I$ is known and $A$ is to be found. We can write this as

$$\tag{3} J(\lambda)=\int\limits_0^\infty dT \ A(T)K(\lambda T) $$

Where $J=\lambda^5 I$, and $K(x)=(e^{1/x}-1)^{-1}$. This integral equation may be solved via Mellin transform. The Mellin transform of $J$ may be related to the transform of $I$

$$\tag{4} J(s)=\int\limits_0^\infty d\lambda \ \lambda^{s-1} J(\lambda)=\int\limits_0^\infty d\lambda \ \lambda^{s-1}\lambda^5 I(\lambda)=I(s+5) $$

The RHS of (3) is in the form of a Mellin convolution, so we have on transforming both sides

$$\tag{5} J(s)=K(s)A(1-s)=I(s+5) $$

Where $J(s)$ and $K(s)$ are the transforms of $J$ and $K$. The Mellin transform of $A$ is now known (presuming all the integrals converge, etc)

$$\tag{6} A(s)=\frac{I(6-s)}{K(1-s)} $$

The Mellin inversion theorem tells us when this may be inverted to recover $A(T)$, which implicitly leads to the conditions on $I$. If it is possible, the formal expression is

$$\tag{7} A(\lambda)=\frac{1}{2\pi i} \int\limits_{c-i\infty}^{c+i\infty} ds \ \lambda^{-s} A(s)=\frac{1}{2\pi i}\int\limits_{c-i\infty}^{c+i\infty} ds \ \lambda^{-s} \frac{I(6-s)}{K(1-s)} $$

It should be noted that the Mellin transform of $K$ looks like it does not have a closed form, and only exists for $\Re(s)<0$.