Show that $\int_0^\infty\frac1{a+(k+(1-x))^2}\:{\rm d}k\le\frac1{\sqrt a}\int_0^{\frac{\sqrt a}{1-x}}\frac1{1+y^2}\:{\rm d}y$

Question

Let $a>0$ and $x\in\mathbb R$. How do we see that $$\int_0^\infty\frac1{a+(k+(1-x))^2}\:{\rm d}k\le\frac1{\sqrt a}\int_0^{\frac{\sqrt a}{1-x}}\frac1{1+y^2}\:{\rm d}y.?$$ This should be just an application of the substitution rule, but I can't figure out how we need to substitute.

Answer 1

Let $$k+1-x=\sqrt{a}t$$ $$\Rightarrow dk=\sqrt{a}dt $$ $$\implies \int_0^\infty\frac{dk}{a+(k+1-x)^2}=\int_\frac{1-x}{\sqrt a}^\infty \frac{\sqrt a dt}{a(1+t^2)}$$

Now apply reciprocal substitution, $y=\frac{1}{x}$ and you are done.

CAVEAT

$1-x$ may be negative, break the integral in that case.