Finding even $f:\mathbb{R} \to \mathbb{R}$ satisfying $\int_{-\infty}^\infty \frac{\arctan x}{xf(x)}dx=1$.

Question

I am trying to find an even, differentiable-everywhere $f:\mathbb{R} \to \mathbb{R}$ such that

$$\int_{-\infty}^\infty \frac{\arctan x}{xf(x)}dx=1.$$

So far, all I have shown is that $f(x)$ is not of the form $x^z$, as the integral would not converge. The only obvious solution I can find that fits the bill is $f(x)=A^{(x^2)}$, though I have no idea how I could isolate $A$ from $$\int_{-\infty}^\infty \frac{\arctan x}{xA^{(x^2)}}dx=1.$$ Numerically, I see $A$ is very close to $17.5$. But I don't see this getting much farther and am wondering if there is a better way to approach this problem.

Answer 1

Since

$$ \int_{-\infty}^\infty\frac{1}{\pi(1+x^2)}\,dx=1 $$

let $$ \frac{\arctan x}{xf(x)}=\frac{1}{\pi(1+x^2)} $$

giving

\begin{eqnarray} f(x)&=&\frac{\pi(1+x^2)\arctan(x)}{x}&&\text{ for }x\ne0\\ &=&\pi&&\text{ for }x=0 \end{eqnarray}

This will be continuous since

$$ \lim_{x\to0}\frac{\pi(1+x^2)\arctan(x)}{x}=\pi $$