# Thread: existence of the definite integral

1. ## existence of the definite integral

I do not know how to proof that integral $\int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx$ exists. I have used Taylor or convergence tests but I do not see the next steps.

2. ## Re: existence of the definite integral

Originally Posted by Camille91
I do not know how to proof that integral $\int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx$ exists. I have used Taylor or convergence tests but I do not see the next steps.
Use the substitution $x-1=\xi$ and then observe that $\lim_{\xi \rightarrow 0}\frac{\ln ^{\pi} (1+\xi)}{\xi^{3}}=0$...

Kind regards

$\chi$ $\sigma$

3. ## Re: existence of the definite integral

I guess now I should do: $\int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx =\underbrace{ \int^{A}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{1} +\underbrace{\int^{\infty}_{A} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{2}$ for $A>1$. With Your observation I conclude the first integral converges but how about the second one?

Thank you for your last post.

4. ## Re: existence of the definite integral

Originally Posted by Camille91
I guess now I should do: $\int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx =\underbrace{ \int^{A}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{1} +\underbrace{\int^{\infty}_{A} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{2}$ for $A>1$. With Your observation I conclude the first integral converges but how about the second one?

Thank you for your last post.
The second integral converges because is...

$\lim_{x \rightarrow \infty} \frac{\ln ^{a} x}{x^{b}} = 0\ ,\ a>0\ , b>0$ (1)

... no matter what are a and b...

Kind regards

$\chi$ $\sigma$

5. ## Re: existence of the definite integral

Using $\lim_{x \rightarrow \infty} \frac{\ln ^{\pi}x}{x^3}=0$ I can prove the integer satisfied necessary condition but it is not sufficient. On the other hand in convergence tests limes to zero say nothing (I mean limit comparison test). Maybe I am wrong?

6. ## Re: existence of the definite integral

Originally Posted by Camille91
Using $\lim_{x \rightarrow \infty} \frac{\ln ^{\pi}x}{x^3}=0$ I can prove the integer satisfied necessary condition but it is not sufficient. On the other hand in convergence tests limes to zero say nothing (I mean limit comparison test). Maybe I am wrong?
What I intended to say is that for x 'large enough' is...

$\frac{\ln^{\pi} x}{x^{3}}< \frac{1}{x^{3-\varepsilon}}$ (1)

... no matter 'how small' is $\varepsilon>0$. Now the integral...

$\int_{a}^{\infty}\frac{dx}{x^{\alpha}}\ ,\ a>0$ (2)

.... converges for all $\alpha>1$, so that also Your integral converges...

Kind regards

$\chi$ $\sigma$

7. ## Re: existence of the definite integral

You are absolutely right and it solved the whole problem. Thank you.