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This question appeared on a Mathematics PhD Preliminary Examination - Real Analysis section.

Let $Q=\{{0 For what values of $a,b$ is the function

$x^ay^b \int_{0}^{\infty}\frac{1}{(x+t)(y^2+t^2)}dt$

bounded on $Q$?

I've tried integrating by parts but it seemed to make the problem more complicated. Ended up needing to further integrate $ln (x+t)$ or $arctan \frac{t}{y}$.

Is the right first step to, instead, bring the $x^ay^b$ into the integral to try to bound the overall integrand?

If someone could give the right approach for such problems, or just a good starting point, that would be great. Thanks.

1 Answers 1

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Note that $\frac{1}{(x+t)(y^2+t^2)}=\frac{1}{x^2+y^2}\left[\frac{1}{x+t}+\frac{x-t}{y^2+t^2}\right].$ Therefore, $\int_0^\infty\frac{1}{(x+t)(y^2+t^2)}dt=\frac{\frac{\pi x}{2y}-\log x+\log y}{x^2+y^2}.$

Claim: The function is bounded on $Q$ if and only if $a>0$, $b\ge 1$ and $a+b>2$.

Proof: The "only if" part is easier. The three inequalities follow from fixing $y\in(0,1)$, fixing $x\in(0,1)$, and letting $x=y$ respectively. To deduce the "if" part, note that for $0\le\lambda\le 1$ and $u,v>0$, $u^\lambda v^{1-\lambda}\le\lambda u+(1-\lambda)v\le \max(\lambda,1-\lambda)(u+v).\quad (*)$ For $(u,v)=(xy,x^{-1}y^3)$ and $(u,v)=(x^2,y^2)$, choosing appropriate $\lambda$(the inequalities of $a,b$ guarantee the existence of $0\le\lambda\le 1$) respectively to apply $(*)$, the conclusion follows.