Good upper bound for $\int_0^1 (1 + 2x)/\sqrt{x + x^3}$

Question

I am trying to obtain an upper and lower estimate for the integral $$I = \int_0^1 \overbrace{\frac{1}{\sqrt{x^2+1}} \left( \frac{1}{\sqrt{x}} + 2\sqrt{x}\right)}^{\Large f(x)}\,\mathrm{d}x,$$ and an approximate value. Eg finding two integrals such that

$$I_l < I < I_u$$

Since the integrand is strictly decreasing on $[0,1]$, a good lower bound is for example $f(1) = 3/\sqrt{2}$. An even better lower estimate can be found by using the taylor expansion of $f$ around $x=1$. For example $$I_l = \int \frac{3}{\sqrt{2}} - \frac{1}{\sqrt{2}}(x-1) + \frac{1}{3\sqrt{2}}(x-1)^2\,\mathrm{d}x = \frac{43}{12\sqrt{2}} $$

But how can one find a good upper estimate for integral? I tried doing $$ I_u = \int_0^1 \frac{1}{\sqrt{x}} + 2\sqrt{x}\,\mathrm{d}x = \frac{10}{3}$$ However this is not a particularly accurate estimate.

Does anyone have any better suggestions for finding a good upper estimate?

Answer 1

Substituting $x:=t^2$ $\>(0\leq t\leq1)$ we obtain $$I=2\int_0^1{1+2t^2\over\sqrt{1+t^4}}\ dt\doteq 2.98808\ .$$ So the challenge would be to prove at least $I<3$.

Since the function $f(t):=(1+t^4)^{-1/2}$ cannot be well approximated by polynomials over the whole range $[0,1]$ we split the integral up into two parts $I_1:=\int_0^{2/3}$ and $I_2:=\int_{2/3}^1$.

The series $${1\over\sqrt{1+u}}=1-{u\over2}+{3u^2\over8}-{5u^3\over16}+\ldots\qquad\bigl(|u|<1\bigr)$$ is alternating; therefore we have $$I_1\leq2\int_0^{2/3}(1+2t^2)\left(1-{t^4\over2}+{3t^8\over8}\right)\ dt ={38399308\over22733865}\doteq1.68908\ .$$ The graph of $f$ has an inflection point at $t=1$, and a precise approximation will have to take care of that. Therefore we approximate $f$ on $\bigl[{2\over3},1\bigr]$ by its Taylor polynomial $g$ at $1$ of third order. This polynomial computes to $$g(t)={1\over\sqrt{2}}(1 + 2 t - 3 t^2 + t^3)\ .$$ We now have to check whether $f(t)\leq g(t)$ is true for ${2\over3}\leq t\leq1$. To this end we compute $$g^2(1-s)\bigl(1+(1-s)^4\bigr)-1 ={s^4\over2}(3 - 6 s - s^2 + 2 s^3 + 4 s^4 - 4 s^5 + s^6)\ .$$ It is easily veryfied that the right side is $\geq0$ for $0\leq s\leq{1\over3}$. Therefore we now have $$I_2\leq \sqrt{2}\int_{2/3}^1 (1+2t^2)(1 + 2 t - 3 t^2 + t^3)\ dt= \sqrt{2}\,{40219\over43740}\doteq1.30037\ .$$ Adding the two partial results up we finally obtain $$I\leq2.98945\ .$$

Answer 2

Probably the best way to find an upper estimate would be to find a straight line lying above the function $$f(x):=\frac{1}{\sqrt{x^{2}+1}}.$$ One such line is $y=-0.36x+1.08$. Using this as an upper bound for $f$ yields the integral upper bound as $$\int<\int_{0}^{1}(-0.36x+1.08)\left(\frac{1}{\sqrt{x}}+2\sqrt{x}\right)\,dx=\frac{384}{125}=3.072.$$ This is a better estimate than before. Another possibility is to take the straight line I've gotten, and lop it off to make it horizontal at $y=1$. This occurs when $x=2/9$. So, let $$g(x)=\begin{cases}1,&\quad 0\le x\le 2/9\\ -0.36x+1.08,&\quad 2/9<x\le 1.\end{cases}.$$ Using this as an upper bound yields $$\int<\frac{384}{125}-\frac{392\sqrt{2}}{10125}\approx 3.017.$$