In the handbook of integral equations, there are provided solutions for two integrals involving functions $f(x)$ and $y(t)$:
Equation 1: Given $$f(x) = \int_{a}^{x} \frac{y(t) \ dt}{\sqrt{x^2 - t^2}},$$ the solution is $$y(t) = \frac{2}{\pi} \frac{d}{dt} \int_{0}^{t} \frac{xf(x) \ dx}{\sqrt{t^2-x^2}}.$$
Equation 2: Given $$f(x) = \int_{x}^{\infty} \frac{y(t) \ dt}{\sqrt{t^2 - x^2}},$$ the solution is $$y(t) = -\frac{2}{\pi} \frac{d}{dt} \int_{t}^{\infty} \frac{xf(x) \ dx}{\sqrt{x^2 - t^2}}.$$
In my case, I have an expression relating the scattered intensity as a function of the orientational distribution function in the form of equation 2, but with a constant $b$ replacing $\infty$ as the upper integral bound:
$$f(x) = \int_{x}^{b} \frac{y(t) \ dt}{\sqrt{t^2 - x^2}}$$
My question is: Am I permitted to use the solution of equation 2 by substituting $\infty$ with $b$, as follows?
$$y(t) = -\frac{2}{\pi} \frac{d}{dt} \int_{t}^{b} \frac{xf(x) \ dx}{\sqrt{x^2 - t^2}}$$
Thank you in advance for your guidance.
Best regards, David
Yes, you are allowed to do that if $y(t)$ vanishes identically for $t\geq b$. In this case, $$ f(x)=\int_{x}^{\infty} \frac{y(t) \, dt}{\sqrt{t^2 - x^2}}= \int_{x}^{b} \frac{y(t) \, dt}{\sqrt{t^2 - x^2}} \tag{1} $$ and $f(x)=0$ for $x\geq b$, hence $$ \int_{t}^{\infty} \frac{xf(x) \, dx}{\sqrt{x^2 - t^2}}= \int_{t}^{b} \frac{xf(x) \, dx}{\sqrt{x^2 - t^2}}. \tag{2} $$