Integral $\ln(1+x^2)/x$ Depending on a Parameter

Question

I want to calculate $F^\prime(t)$ of the function $$ F(t)=\int_{0}^1 \frac{\ln(1+tx^2)}{x} dx $$ and I know the steps I have to show. One step is showing that $$ F(t)\leq g(x) $$ for some majorant $g$. But I have trouble finding that majorant. I think that it needs to exist. Any hints? Thanks in advance.

Answer 1

Since both $\dfrac{1}{x}$ and $\text{ln}(1+tx^2)$ are continuous in $(0, 1)$ for both $x$ and $t$, clearly $\dfrac{\text{ln}(1+tx^2)}{x}$ is also continuous in $(0,1)$. The logarithm will be continuous with respect to $t$ as long as $t>-1$. We can then the differentiate under the integral sign to obtain: $$ \dfrac{dF(x)}{dt} = \int_0^1 \partial_t\left(\dfrac{\text{ln}(1+tx^2)}{x}\right)dx$$ $$ = \int_0^1 \dfrac{x^2}{x(1+tx^2)}dx = \int_0^1 \dfrac{x}{1+tx^2}dx$$ Let $u=1+tx^2$, we have $du = 2txdx$; $u(0) = 1$, $u(1) = 1+t$: $$\int_0^1 \dfrac{x}{1+tx^2}dx=\dfrac{1}{2t} \int_{1}^{1+t} \dfrac{du}{u} $$ $$= \dfrac{1}{2t}\;ln(1+t) $$