How to show that the integral is convergent without actually evaluating it?

Question

$$ int _ 0 ^ { + infty } frac 1 { sqrt { 1 + x ^ 8 } }  mathrm d x $$

I have a hint that says there is no elementary way to evaluate this integral. So I am supposed to bound this integrand by a simpler function and use the following fact without proof:

Let $ f : [a,+infty) to mathbb R $ be a bounded and increasing function. Then $ lim _ { x to infty } f ( x ) $ exists.

I specifically don't understand what it means to bound the integrand by a simpler function. Any help will be appreciated.

PierreCarre · Accepted Answer

Start by noting that the existence of the proposed integral is equivalent to the existence of the integral $int_1^{+infty}frac{1}{sqrt{1+x^8}} dx$. Then, since
$$
frac{1}{sqrt{1+x^8}} leq frac{1}{sqrt{x^8}} = frac{1}{x^4},
$$
the convergence of $int_1^{+infty} frac{1}{x^4} dx$ implies the convergence of our integral.

Barry Cipra · Answer

Just to take a slightly different approach, if we can establish that
$${1oversqrt{1+x^8}}le{2over1+x^2}$$
for all $x$, then we have
$$int_0^infty{dxoversqrt{1+x^8}}le2int_0^infty{dxover1+x^2}=pi$$
To establish the key inequality, note that
$${1oversqrt{1+x^8}}le{2over1+x^2}iff1+2x^2+x^4le4+4x^8iff 4x^8-x^4-2x^2+3ge0$$
and
$$4x^8-x^4-2x^2+3ge
begin{cases}
0-1-2+3=0quad&text{if }|x|le1\
4x^8-x^8-2x^8+3=x^8+3ge0quad&text{if }|x|ge1
end{cases}$$
The only advantage (if any) of this approach is that it uses a single formula for the bounding function, rather than one formula for $xle1$ and another for $xge1$. But doing so, of course, comes at the expense of a slightly tricky verification of a polynomial inequality.

Eric Towers · Answer

For $x > 1$, begin{align*} 0 < x^8 &< x^8 + 1 < infty \ 0 < x^4 = sqrt{x^8} &< sqrt{x^8 + 1} < infty \ infty > frac{1}{x^4} &> frac{1}{sqrt{x^8+1}} > 0 text{.} end{align*} So, $$ 0 = int_1^infty 0 ,mathrm{d}x < int_1^infty frac{1}{sqrt{x^8 + 1}} ,mathrm{d}x < int_1^infty frac{1}{x^4} ,mathrm{d}x text{.} $$ The right-hand $p$-integral is convergent, so $displaystyle int_1^infty frac{1}{sqrt{x^8 + 1}} ,mathrm{d}x$ is convergent.

For $0 leq x leq 1$, begin{align*} 1 &leq x^8+1 leq 2 \ 1 = sqrt{1} &leq sqrt{x^8+1} leq sqrt{2} \ 1 = frac{1}{1} &geq frac{1}{sqrt{x^8+1}} geq frac{1}{sqrt{2}} text{.} \ end{align*} So, $$ 1 = int_0^1 1 ,mathrm{d}x geq int_0^1 frac{1}{sqrt{x^8+1}} ,mathrm{d}x geq int_0^1 frac{1}{sqrt{2}} ,mathrm{d}x = frac{1}{sqrt{2}} text{.} $$ This shows the integral $displaystyle int_0^1 frac{1}{sqrt{x^8+1}} ,mathrm{d}x $ is convergent.

Therefore, $displaystyle int_0^infty frac{1}{sqrt{x^8+1}} ,mathrm{d}x$ is convergent.

zkutch · Answer

Hint: $$ frac{1} { sqrt{1+ x^{8} } } sim frac{1} { x^4 }, x to +infty$$

How to show that the integral is convergent without actually evaluating it?

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