The answer is $\boxed{a=b}$. This can be shown by using the big-O convention.

Because $(1+x)^{\frac{1}{2}}=1+\frac{x}{2}+O(x^2)$,  hence we have $$\sqrt{x+a}-\sqrt{x}=x^{\frac{1}{2}}\left(\sqrt{1+\frac{a}{x}}-1\right)=x^{\frac{1}{2}}\left(1+\frac{a}{2x}+O(x^{-2})-1\right)$$

It follows that $$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{\frac{1}{4}}\left(\frac{a}{4x}+O(x^{-2})\right)$$

and $$\sqrt{\sqrt{x+b}-\sqrt{x}}=x^{\frac{1}{4}}\left(\frac{b}{4x}+O(x^{-2})\right)$$

It follows that $$\int_b^{\infty}\left(\sqrt{\sqrt{x+a}-\sqrt{x}}-\sqrt{\sqrt{x}-\sqrt{x-b}}\right)dx=\int_b^{\infty}x^{-\frac{1}{4}}\left(\frac{a-b}{4x}+O(x^{-2})\right)dx=\int_b^{\infty}\left(\frac{a-b}{4}x^{-\frac{3}{4}}+O\left(x^{-\frac{7}{4}}\right)\right)dx$$

The second term above is a constant multiple of $x^{-\frac{7}{4}}$, hence converging. However, the first term will diverge unless $a=b$ which eliminates it. Hence, the final answer is $a=b$.