# Finding the global minimum of $\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}\,dx$ having just the local minimum.

In order to calculate the values of $$a$$ and $$b$$ such we get the minimum possible for:

$$\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}\,dx$$

I got the help of @TheSimpliFire among others to get the respectively $$a$$ and $$b$$ here:

Find $a$ and $b$ for which $\int_{0}^{1}( ax+b+\frac{1}{1+x^{2}} )^{2}\,dx$ takes its minimum possible value.

Then, as we found were the partial derivatives of $$a$$ and $$b$$ are zero it is not hard to prove the founded $$(a,b)$$ satisfies that for:

$$D(a,b) = f_{xx}'(a,b)f_{yy}'(a,b)-f_{xy}(a,b)^2$$

As for $$p=(a,b)$$;$$D(p) > 0 \land f_{xx}'(p) > 0 \implies \text{minimum}$$ i s satisfied, then $$(a,b)$$ are locally minimum. My question is how to verify $$(a,b)$$ is also the global minimum?? Thanks!!!

Since $$F=\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}\,dx=\frac{a^2}{3}+a (b+\log (2))+\frac{1}{8} \left(8 b^2+4 \pi b+\pi +2\right)$$ $$\frac{\partial F}{\partial a}=\frac{2 a}{3}+b+\log (2)=0 \implies a=-\frac{3}{2} (b+\log (2))$$ Replace in $$F$$ to get $$F=\frac{1}{8} \left(2+\pi +6 \pi ^2+48 \log ^2(2)-36 \pi \log (2)\right)+ (2 \pi -6 \log (2))b+b^2$$ which is just a quadratic in $$b$$ and shows a single minimum for $$b=3\log(2)-\pi \implies a=\frac{3}{2} (\pi -4 \log (2))$$ $$\implies F_{min}=\frac{1}{8} \left(2+\pi -2 \pi ^2-24 \log ^2(2)+12 \pi \log (2)\right)$$
Call your integral $$f(a,b)$$. You have found a single point $${\bf p}\in{\mathbb R}^2$$ with $$\nabla f({\bf p})={\bf 0}$$. Let $$f({\bf p})=:c>0$$.
Inspecting the expression defining $$f(a,b)$$ it is obvious that there is an $$M>0$$ such that $$f(a,b)\geq 2c$$ when $$a^2+b^2\geq M^2$$. The disc $$B_M: \>a^2+b^2\leq M^2$$ is compact, and $$f$$ is continuous on $$B_M$$. Therefore $$f$$ assumes a minimum $$\leq c$$ on $$B_M$$, and this cannot be at a point $${\bf q}\in\partial B_M$$. This minimum is therefore a local minimum of $$f$$ at an interior point of $$B_M$$, hence $$\nabla f({\bf q})={\bf 0}$$ at this point. There is only one such point, namey the point $${\bf p}$$ found in your calculations.