I've got the following question: We've got an Abel integral equation of the first kind $$\int\limits_0^t {\frac{{f(\tau )}}{{\sqrt {t - \tau } }}d\tau } = \varphi (t)$$ and the solution has got the form $$f(t) = \frac{1}{\pi }\frac{d}{{dt}}\int\limits_0^t {\frac{{\varphi (\tau )dt}}{{\sqrt {t - \tau } }}} = \frac{{\varphi (0 + )}}{{\pi \sqrt t }} + \frac{1}{\pi }\int\limits_0^t {\frac{{\varphi '(\tau )}}{{\sqrt {t - \tau } }}d\tau } $$ but, what if $$\mathop {\lim }\limits_{t \to 0 + } \varphi (t) = + \infty $$ Is this formula true? I.e. for $\varphi (t) = - \ln (t)$ we get $$f(t) = - \frac{{\ln (4t)}}{{\pi \sqrt t }}$$ I get it using first part (integration and differentiation after). But I thought we can differentiate under integral sign because $\frac{{\varphi (t)}}{{\sqrt {t - \tau } }}$ is continious in $[0,t]$.
Actually I'm dealing with $$\int\limits_0^t {{{f(\tau ){e^{\lambda (t - \tau )}}} \over {\sqrt {t - \tau } }}d\tau } = \varphi (t)$$ and I want to find the behaviour $f(t)$ at $t \to +0$ if I know $\varphi (t)$. Thanks in advance!