Solve the following Fredholm Integral Equation

Question

Solve the Integral Equation :$$y(x)=\frac{6}{5}(1-4x)+\lambda\int_0^1(x\ln t-t\ln x)y(t)\,dt$$

Let , $$y(x)=\frac{6}{5}(1-4x)+\lambda xC_1-\lambda\ln x C_2$$where,

$$C_1=\int_0^1\ln t\left[\frac{6}{5}(1-4t)+\lambda C_1 t-\lambda C_2 \ln t\right]\,dt$$

My problem is on the integration $\int_0^1t\ln t\,dt$. The integral is improper at $t=0$ and the integral is NOT convergent....

How we solve the problem ??

Answer 1

$$ \int_0 ^1 t \ln t \, dt = \frac{t^2}{2}(\ln t - \frac{1}{2}) \big|_{\alpha \to 0^+} ^{ 1 } = \frac{-1}{4} - \lim_{ \alpha \to 0^+} ( \frac{ \ln \alpha - \frac{1}{2} }{ 1 / (\alpha ^2 / 2) } ) . $$ Note that the limit is indeterminate, as it goes to $\frac{\infty}{\infty}$, and so we apply L'Hospital's rule to calculate $$ \int_0 ^1 t \ln t \, dt = - \frac{1}{4} - \frac{1}{2} \lim_{ \alpha \to 0^+} \frac{ \frac{1}{\alpha} }{ -2 \alpha^{-3} } = - \frac{1}{4} + \frac{1}{4} \cdot \lim_{ \alpha \to 0^+} \alpha^2 = - \frac{1}{4} . $$

Answer 2

Consider $$I=\int t \log(t)\,dt$$ Integrate by parts using $u=\log(t)$, $v'=t\,dt$ which give $u'=\frac {dt}t$, $v=\frac{t^2}2$. So $$I=\int t \log(t)\,dt=\frac{1}{2} t^2 \log (t)-\frac{t^2}{4}$$ which does not make any problem even for $t=0$.