# Find $f(x)$ such that $\int_{-\infty}^\infty \int_{-a}^a \frac{|y+x|^p}{1+y^2} f(x) dx dy<\infty$

Can we find a non-negative function $f(x)$ on $[-a,a]$ such that \begin{align} \int_{-\infty}^\infty \int_{-a}^a \frac{|y+x|^p}{1+y^2} f(x) dx dy <\infty \end{align} and can be computed in closed form?

Note we must restrict ourselves to $p \in [0,1)$ otherwise the integral is infinity.

I tried and example when $f(x)$ is a constant (which I thought would be the simplest) but could not find a close form expression integrate.

• Actually you need $p<1.$ – zhw. Nov 22 '16 at 22:30
• @zhw. Yes, you right. Will Correct. Any idea how to find such a function $f$? – Boby Nov 22 '16 at 22:51

Any $f$ integrable on $[-a,a]$ will give convergence for $0\le p <1.$ That's because $|y+x|^p \le |y|^p + |x|^p.$ It then follows that your integral is no more than
\begin{align} \int_{-\infty}^\infty \int_{-a}^a \frac{|y|^p+|x|^p}{1+y^2}\, f(x)\, dx\, dy = \int_{-\infty}^\infty \int_{-a}^a \frac{|y|^p}{1+y^2} \, f(x)\, dx\, dy + \int_{-\infty}^\infty \int_{-a}^a \frac{|x|^p}{1+y^2} \, f(x)\, dx\, dy \end{align},