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

Calculate for which values $$a$$ and $$b$$ the integral

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

takes its minimum possible value?

For being honest I'm not sure how to try this, but my idea is to calculate its derivative using fundamental calculus theorem as $$\left(ax+b+\frac{1}{1+x^{2}} \right)^{2}$$ is a continuous function over $$[0,1]$$. And then, evaluate the integral over $$0,1$$ and the values where the derivative we calculate is zero and find which $$a$$ and $$b$$ does the work. Sorry but this is the first problem of this type I'm trying. Thanks

• Actually, you can expand the integrand. After squaring, all the integrals are elementary, so you can calculate them and obtain the function $f(a,b)$ to be minimized explicitly. – A.Γ. Apr 20 at 18:28
• Let $f(a,b) = \int_0^1(ax + b + \frac{1}{1 + x^2})dx$. Integrate on x and then minimize the function on the real plane (since you have no boundary condition, minimization is quite easy) – Francisco José Letterio Apr 20 at 18:30
• What range of $a$ and $b$ are you considering? If $a,b \geq 0$, then obviously $a=b=0$ minimizes.. – Dzoooks Apr 20 at 18:38
• $a$ and $b$ where given without any restrictions. @Dzoooks – Cos Apr 20 at 18:40

Let \begin{align}f(a,b)=\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}dx\implies \frac{\partial f(a,b)}{\partial a}&=\int_0^12x\left( ax+b+\frac{1}{1+x^{2}} \right)\,dx\\&=\left[\frac{2ax^3}3+bx^2+\ln(1+x^2)\right]_0^1\end{align} so $$\frac{\partial f(a,b)}{\partial a}=\frac23a+b+\ln2=0\tag1$$ for critical points. Similarly \begin{align}f(a,b)=\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}dx\implies \frac{\partial f(a,b)}{\partial b}&=\int_0^12\left( ax+b+\frac{1}{1+x^{2}} \right)\,dx\\&=\left[ax^2+2bx+2\tan^{-1}x\right]_0^1\end{align} so $$\frac{\partial f(a,b)}{\partial b}=a+2b+\frac\pi2=0\implies\frac12a+b+\frac\pi4=0\tag2$$ for critical points. Performing $$(1)-(2)$$ gives $$\frac16a=\frac\pi4-\ln2\implies a=\boxed{\frac{3\pi}2-6\ln2}$$ and putting this into $$(2)$$ gives $$b=-\frac12a-\frac\pi4=\boxed{-\pi+3\ln2}.$$

• Thanks! But Im not really sure why$2x$ appears from when yoy calculate $\frac{\partial f(a,b)}{\partial a}$ ?? @TheSimpliFire – Cos Apr 20 at 19:11
• @Cos Since $\frac{\partial}{\partial a}\left(ax+b+\frac{1}{1+x^2}\right)^2 = 2\left(ax+b+\frac{1}{1+x^2}\right)\cdot x$ by the chain rule – TheSimpliFire Apr 20 at 19:14
• Thanks for the help @TheSimpliFire – user1952500 Apr 20 at 19:20
• No worries @user1952500, mistakes are a good learning experience – TheSimpliFire Apr 20 at 19:20

We can also solve this problem by projecting $$f(x) = -\dfrac{1}{1+x^2}$$ on the linear space spanned by all linear functions defined on $$[0,1]$$ where we use the inner product $$\langle f,g\rangle=\int_0^1f(x) g(x) dx$$.

The normalized constant function $$e_0(x) = 1$$ can be taken to be one basis vector of the linear space of linear function. The function $$h(x) = x$$ is linearly independent from $$e_0(x)$$ but it is not orthogonal to it. Using the Gram–Schmidt process we can find the correct basis vector as follows. We subtract from $$h(x)$$ its component in the direction of $$e_0(x)$$ and then we normalize the result. We put:

$$g(x) = h(x) - \langle h,e_0\rangle e_0(x) = x - \int_0^1 x dx = x - \frac{1}{2}$$

Normalizing $$g(x)$$ then gives us the other basis vector $$e_1(x)$$ of the space of the linear functions:

$$e_1(x) = \frac{g(x)}{\sqrt{\langle g,g\rangle}} = \frac{x-\frac{1}{2}}{\sqrt{\int_0^1 \left(x-\frac{1}{2}\right)^2 dx }} = 2\sqrt{3}\left(x-\frac{1}{2}\right)$$

The projection of $$f(x)$$ on the linear space spanned by the linear functions is then:

$$\langle f,e_0 \rangle e_0(x) + \langle f,e_1 \rangle e_1(x) = 3\log(2) -\pi +\left( \frac{3\pi}{2}-6\log(2)\right)x$$