# Solutions of Fredholm integral equation of the fisrt kind with asymmetric kernel

Given the integral equation: $$\int_0^{\lambda_0}K(\lambda,T)S(\lambda)d\lambda=f(T)$$ with $S(\lambda)$ unknown function and the kernel: $$K(\lambda,T)=\frac{1}{\lambda^5(\exp(k/\lambda T )-1)}$$ with $f(T)$ continuous function: $0\le f(T)\lt +\infty$ for $0\lt T\lt T_0$, I'm looking for a solution of this equation. Because I suppose there is no smple way to solve it analytically, I tried to solve it numerically. So, because I know from 'a priori' information the $S(\lambda)$ can expanded in series in $0\le\lambda\le\lambda_0$as: $$S(\lambda)=\sum_{k=0}^Na_k\lambda^k$$ I wrote the previous equation as: $$\sum_{k=1}^Na_k\int_0^{\lambda_0}d\lambda K(\lambda,T)\lambda^k=f(T)$$ and then: $$f(T_1)=\sum_{k=1}^Na_kg_1(k)$$ with $g_1(k)=\int_0^{\lambda_0}d\lambda K(\lambda,T_1)\lambda^k$ and in general: $$f(T_j)=\sum_{k=1}^Na_kg_j(k)$$ with $1\le j\le N$ solving the system in $a_k$ I can find an approximate solution of the integral equation. Is this simple method correct?

• For motivation, you might discuss the appearance of the Planck distribution as a kernel. – Steve Huntsman Oct 25 '13 at 10:13
• @SteveHuntsman: the integral equation can be obtained from a method useful to estimate the spectral response of a photodetector. – Riccardo.Alestra Oct 25 '13 at 11:15
• Actually if you divide both sides by $T^4$ and change all variables to their logarithms, you'll get a convolution equation, after which you'll have access to all the associated machinery. – fedja Oct 26 '13 at 2:01