# Convergence of the integral $\int_0^\infty \frac {2x+14}{x^3+1}dx$

Theorem $1:$
Suppose $0\leq f(x)\leq g(x)$ for all sufficiently large $x$.
If $\int_a^\infty g(x)dx$ converges, so does $\int_a^\infty f(x)dx$
If $\int_a^\infty f(x)dx$ diverges, so does $\int_a^\infty g(x)dx$

Theorem $2:$ If $0\leq f(x)\leq Cx^{-p}$ for all sufficiently large $x$, where $p\gt 1$, then $\int_a^\infty f(x)dx$ converges.

Note also that $\int_1^\infty x^{-p}dx$ converges iff $p\gt 1$

Now, in the book (using theorem $2$) it says the integral $$\int_0^\infty \frac {2x+14}{x^3+1}dx$$ converges because $$\frac {2x+14}{x^3+1}\leq \frac {4x}{x^3}=\frac {4}{x^2} , \qquad for \quad x\geq 7$$ What if I wanted to use theorem $1$? I know that $\int_1^\infty x^{-2}dx$ converges and $\frac {2x+14}{x^3+1}\leq {4}{x^{-2}} , \qquad for \quad x\geq 7$

However in theorem $1$ it says that "If $\int_a^\infty g(x)dx$ converges, so does $\int_a^\infty f(x)dx$": Here the lower bounds of the integrals are the same, but in our example the lower bound of the integral is $0$ in $\int_0^\infty \frac {2x+14}{x^3+1}dx$ and it is $1$ in $\int_1^\infty x^{-2}dx$. Doesn't this make any difference?
I also saw a remark which is:

If $c\gt a$, the convergence of $\int_a^\infty f(x)dx$ is equivalent to that of $\int_c^\infty f(x)dx$

So can we say that the convergence of $\int_0^\infty \frac {2x+14}{x^3+1}dx$ is equivalent to the convergence of $\int_1^\infty \frac {2x+14}{x^3+1}dx$ so that now theorem $1$ can be applied correctly?

HINT

Note that

$$\int_0^\infty \frac {2x+14}{x^3+1}dx=\int_0^1\frac {2x+14}{x^3+1}dx+\int_1^\infty \frac {2x+14}{x^3+1}dx$$

and the first is a proper integral; for the second note that

$$\frac {2x+14}{x^3+1}\sim \frac1{x^2}$$

and use limit comparison test.

• Thank you! What about the question that I've asked after the remark? – Leyla Alkan Mar 5 '18 at 15:57
• @LeylaAlkan Yes of course, the "problem" is at $\infty$ not at $x=0$, thus the convergence can be evaluated for any interval $(a, \infty)$ with $a>0$. – gimusi Mar 5 '18 at 15:59

Now $\displaystyle\int_{7}^{\infty}\dfrac{2x+14}{x^{3}+1}dx<\infty$ as with the same "lower bound" that you have indicated.

But $\dfrac{2x+14}{x^{3}+1}$ is continuous on $[0,7]$, so $\displaystyle\int_{0}^{7}\dfrac{2x+4}{x^{3}+1}dx<\infty$, so summing up these two integrals, we have $\displaystyle\int_{0}^{\infty}\dfrac{2x+4}{x^{3}+1}dx=\displaystyle\int_{0}^{7}\dfrac{2x+4}{x^{3}+1}dx+\displaystyle\int_{7}^{\infty}\dfrac{2x+4}{x^{3}+1}dx<\infty$.