 Solution

The product $(\tanh s)y(s)$ is continuous, so the fundamental theorem of calculus makes the integral continuously differentiable. Differentiating the integral equation and then evaluating at zero gives $y^{\prime }=(\tanh x)y$, $y(0)=1$. Since $(\log \cosh x{)}^{\prime }=\tanh x$, an integrating factor is $1/\cosh x$. Thus $(y/\cosh x{)}^{\prime }=0$, andPut $L(x)=\log \cosh x$, so $L(0)=0$ and $L^{\prime }=\tanh x$. The claimed Picard iterate formula holds for $n=0$. Assuming it at $n$, termwise integration of the finite sum givesThis proves by inductionThe same calculation works for negative $x$ because it uses an antiderivative rather than a monotonicity substitution. On every bounded interval $|x|\le R$, $0\le L(x)\le \log \cosh R$, so comparison with the exponential-series tail also proves uniform convergence there.