# improper integrals (comparison theorem)

In my assignment I have to evaluate the (improper) integral, by means of the "comparison theorem". And note whether the function is divergent or convergent. $$\int^{\infty}_{0} \frac{x}{x^3 + 1}dx$$

The comparison theorem basically says

Suppose $f$ and $g$ are continuous functions with $f(x) \geq (x)$ for $x \geq a$. Then:
A) if $\int^\infty_af(x)dx$ is convergent then $\int^\infty_ag(x)dx$ is convergent
B) if $\int^\infty_ag(x)dx$ is divergent then $\int^\infty_af(x)dx$ is divergent

So in other words: to prove if a given integral is convergent you find a function whose integral is larger than the given integral (within the boundaries). And to prove divergence you find a "divergent integral" whose function is always smaller than the function is question.

Now how should I go with this? Is there any trick to the above? Can I "see" (without calculator/automatic plotting) if an integral will be divergent or convergent (so to reduce time)?

I think $$\int_0^\infty 1/x^2$$ diverges because ,in $[0,1]$ given integral diverges. What we have to do is split the given integral like this. $$\int_0^\infty \frac {x}{x^3+1}= \int_0^1 \frac {x}{x^3+1} + \int_1^\infty \frac {x}{x^3+1}$$ Definitely second integral converges. Taking first integral We have $$x\leq x^4$$ for $x\in [0,1]$ So given function $$\frac {x}{x^3+1} \leq \frac {x^4}{x^3+1} \leq \frac {x^4}{x^3} = x$$ Since $g(x)=x$ is convegent in $[0,1]$, first integral convergent Hence given integral converges

You can "see" the convergence of the integral by noting that assimptotically $\frac{x}{x^3+1}\sim \frac{1}{x^2}$. As $\int_a^{\infty}\frac{1}{x^2}dx$ converges when $a>0$, so does $\int_a^{\infty}\frac{x}{x^3+1}dx$. This argument can be made precise by using the fact that for $x>1$, $\frac{x}{x^3+1}=\frac{1}{x^2+\frac{1}{x}}<\frac{1}{x^2}$, and applying the comparison theorem you stated.

Hint: $$\frac { x }{ { x }^{ 3 }+1 } <\frac { x }{ { x }^{ 3 } } =\frac { 1 }{ { x }^{ 2 } }$$ for $x>0$

As far as I know $\int_{0}^{\infty} \frac{1}{x^2}dx$ diverges so probably the hints above are not good since you want to verify the convergence of $$\int_0^{\infty} \frac{x}{x^3+1}dx.$$ I'm not sure whether it is the best evaluation but e.g. since $x^3+1 = (x+1)(x^2-x+1)$ we have $\frac{x}{x^3+1} < \frac{x+1}{x^3+1} = \frac{1}{x^2-x+1}$

and we can show that $\int_0^{\infty} \frac{1}{x^2-x+1}dx = \frac{4\sqrt{3}\pi}{9}$ so your integral $\int_0^\infty \frac{x}{x^3+1}$ converges. Although, there might be a better evaluation which gives you ` a better' function $g(x) > \frac{x}{x^3+1}$ whose integral is trivial to solve.

• $int^\infty_0(1/x^p)dx$ converges if p > 1.... Oct 21, 2013 at 13:43
• $\int_{0}^{\infty} \frac{1}{x^2} dx = \left[-\frac{1}{x}\right]_{0}^{\infty} = \lim_{x \to 0} \frac{1}{x} = \infty$ Oct 21, 2013 at 13:46
• If you don't believe above calculations maybe you will believe the computer wolframalpha.com/input/… $\sum \frac{1}{n^p}$ converges if $p>1$ it is different with the integrals. Oct 21, 2013 at 13:51
• But how does the integrand behave near the origin? more like $x$. The integral converges. Oct 21, 2013 at 14:00
• $\int_{\frac{1}{n}}^{\infty} \frac{1}{x^2}dx = n$ it converges if $n > 0$. I don't understand your comment. Oct 21, 2013 at 14:04