Improper integral and the ordinary differential equation 2

Question

How to prove that

$$f(t)=\int_0^{\infty}\frac{e^{-tx}}{1+x^2}dx$$

satisfies

$$f''(t)+f(t)=\frac1t ?$$

Answer 1

We use the same method used here:

Let $$h(t,x)=\frac{e^{-tx}}{1+x^2}$$ then $$\frac{\partial h}{\partial t}(t,x)=-\frac{xe^{-tx}}{1+x^2}$$ and

$$\frac{\partial^2 h}{\partial t^2}(t,x)=\frac{x^2e^{-tx}}{1+x^2}$$ and we have $$\left|\frac{\partial h}{\partial t}(t,x)\right|\le\frac{xe^{-ax}}{1+x^2}\in L^1([a,+\infty)),\;\forall a>0$$ and $$\left|\frac{\partial^2 h}{\partial t^2}(t,x)\right|\le\frac{x^2e^{-ax}}{1+x^2}\in L^1([a,+\infty)),\;\forall a>0$$ hence $y$ is twice differentiable on $(0,+\infty)$ (in fact $C^\infty(0,+\infty)$) and $$y''(t)=\int_0^\infty\frac{x^2e^{-tx}}{1+x^2}dx$$

Now from this expression of $y''$ we write $$y''(t)=\int_0^\infty\frac{x^2e^{-tx}}{1+x^2}dx=\int_0^\infty\frac{(x^2+1-1)e^{-tx}}{1+x^2}dx=\int_0^\infty{e^{-tx}}dx-y\\=-\frac1te^{-tx}\bigg|_0^\infty-y=\frac1t-y$$ so we conclude $$y''+y=\frac1t$$