1

Probably just dont see it, was never good with prooving existance.

Prove that this integral exists

$$\int_0^4dx \int_{\sqrt x}^{min[2,2\sqrt x]} \frac{dy}{\sqrt {x+y^2}}$$

Then calcualte.

I would start with $$\int_0^4dx \int_{\sqrt x}^{min[2,2\sqrt x]} \frac{dy}{\sqrt {x+y^2}}=\int_0^1dx \int_{\sqrt x}^{2\sqrt x} \frac{dy}{\sqrt {x+y^2}}+\int_1^4dx \int_{\sqrt x}^{2} \frac{dy}{\sqrt {x+y^2}}$$

Where I find at both cases the integration limit with $\sqrt x$ to be the problematic one. But as said and not sure and tend to get stuck at these.

Neither can do I see a good substitution to use for the later calculation.

Katpton Liamfuppinshire
  • 317

1 Answers1

2

Substitution $y = \sqrt{x}u$ gives

$$ \int_{0}^{4} dx \int_{\sqrt{x}}^{2\wedge 2\sqrt{x}} \frac{dy}{\sqrt{x+y^2}} = \int_{0}^{4} dx \int_{1}^{\frac{2}{\sqrt{x}} \wedge 2} \frac{du}{\sqrt{1+u^2}}. $$

Now from the simple estimate

$$ \int_{1}^{\frac{2}{\sqrt{x}} \wedge 2} \frac{du}{\sqrt{1+u^2}} \leq \int_{1}^{2} \frac{du}{\sqrt{1+u^2}} < \int_{1}^{2} \frac{du}{u} = \log 2, $$

we find that the integral is bounded above by $4\log 2 < \infty$.

Sangchul Lee
  • 167,468