# Thread: How to determine the convergence

1. ## How to determine the convergence

Hi, I don't know to determine the convergence of these integrals:

The first one is $\displaystyle {\displaystyle \int_{0}^{1}\frac{3\sin^{2}(x)+5\cos^{5}(x)}{\sqrt {x}(x+1)}}dx$

and the second is $\displaystyle {\displaystyle \int_{1}^{\infty}\frac{2+x^{3}}{1+x^{6}}dx}$

$\displaystyle \displaystyle$

For the second I tried the following $\displaystyle {\displaystyle I=\int_{1}^{\infty}\frac{2}{x^{6}(1+\frac{1}{x^{6} })}dx}+\int_{1}^{\infty}\frac{1}{x^{3}(1+\frac{1}{ x^{6}})}dx=\int_{1}^{\infty}f_{1}(x)dx+\int_{1}^{\ infty}f_{2}(x)dx$
When $\displaystyle \displaystyle x\rightarrow\infty \begin{cases} x^{6}\rightarrow & \infty\\ (1+\frac{1}{x^{6}}) & \rightarrow1\\ x^{3} & \rightarrow\infty\end{cases}$

Then
$\displaystyle f_{1}(x)\sim\frac{2}{x^{6}}$ , Thus$\displaystyle I_{1}$converges

$\displaystyle {\displaystyle f_{2}(x)}\sim\frac{1}{x_{3}}\Rightarrow$ $\displaystyle I_{2}$ converges, then$\displaystyle I$ converges.
my reasoning is correct?

Bye and thanks for everything.

2. No, that piece-meal convergence idea is not as useful as you want it to be. That's why there is no such theorem. You just need to make sense of it. You're just demonstrating convergence.

$\displaystyle \frac{2+x^{3}}{1+x^{6}}\;<\;\frac{2+x^{3}}{x^{6}}$

Now, it's pretty obvious. I just made the denominator a little smaller.

Let's just play with the other one and see if it leads anywhere...

With only a little thought:

$\displaystyle \frac{-\sqrt{34}}{\sqrt{x}(x+1)}\;<\;\frac{3\sin^{2}(x) + 5\cos^{5}(x)}{\sqrt{x}(x+1)}\;<\;\frac{\sqrt{34}}{ \sqrt{x}(x+1)}$

We could narrow it down more than that, but that may suffice.

$\displaystyle \frac{-\sqrt{34}}{\sqrt{x+1}(x+1)}\;<\;\frac{-\sqrt{34}}{\sqrt{x}(x+1)}\;<\;\frac{3\sin^{2}(x) + 5\cos^{5}(x)}{\sqrt{x}(x+1)}\;<\;\frac{\sqrt{34}}{ \sqrt{x}(x+1)}\;<\;\frac{\sqrt{34}}{\sqrt{x}(x)}$

What say you? Did that get us anywhere? Actually, no, but let's see what you can present.