# Integration goes wrong? ( ln(x)/x )

• Mar 10th 2011, 10:00 AM
Pranas
Integration goes wrong? ( ln(x)/x )
$\displaystyle \displaystyle $I = \int\limits_0^{ + \infty } {\frac{{\ln x}}{x}} dx = \mathop {\lim }\limits_{\gamma \to + \infty } \left( {\int\limits_{1/\gamma }^\gamma {\frac{{\ln x}}{x}dx} } \right)$$

Because $\displaystyle \displaystyle $\int {\frac{{\ln x}}{x}dx = \int {\ln x \cdot d\left( {\ln x} \right)} } = \frac{{{{\left( {\ln x} \right)}^2}}}{2}$$

$\displaystyle \displaystyle $I = \frac{1}{2}\mathop {\lim }\limits_{\gamma \to + \infty } \left( {{{\left( {\ln \gamma } \right)}^2} - {{\left( {\ln \frac{1}{\gamma }} \right)}^2}} \right) = \frac{1}{2}\mathop {\lim }\limits_{\gamma \to + \infty } \left( {{{\left( {\ln \gamma } \right)}^2} - {{\left( { - \ln \gamma } \right)}^2}} \right)$$

So it seems like zero, bus this feels kind of wrong to me. Besides, computer software assumes that integral diverges.
What should I do?
• Mar 10th 2011, 10:06 AM
Ackbeet
Couple of things:

1. $\displaystyle \infty-\infty$ is undefined. It's like dividing by zero: not allowed.

2. Your integral most definitely diverges on the positive end. Examining the integral

$\displaystyle \displaystyle\int_{e}^{\infty}\frac{\ln(x)}{x}\,dx \ge \int_{e}^{\infty}\frac{1}{x}\,dx,$

you see that it diverges. I'd have to think a bit more about what happens at the lower limit.
• Mar 10th 2011, 10:23 AM
Pranas
Quote:

Originally Posted by Ackbeet
Couple of things:

1. $\displaystyle \infty-\infty$ is undefined. It's like dividing by zero: not allowed.

2. Your integral most definitely diverges on the positive end. Examining the integral

$\displaystyle \displaystyle\int_{e}^{\infty}\frac{\ln(x)}{x}\,dx \ge \int_{e}^{\infty}\frac{1}{x}\,dx,$

you see that it diverges. I'd have to think a bit more about what happens at the lower limit.

Somewhat of a similar behaviour:

$\displaystyle \displaystyle $\int\limits_{ - \pi /2}^{\pi /2} {\tan x} \cdot dx = 0$$

I find no other way but to accept this by definition even though there's like $\displaystyle \infty-\infty$ involved.

I'm afraid situation here might be relatively similar :(
• Mar 10th 2011, 10:27 AM
Ackbeet
Quote:

Originally Posted by Pranas
Somewhat of a similar behaviour:

$\displaystyle \displaystyle $\int\limits_{ - \pi /2}^{\pi /2} {\tan x} \cdot dx = 0$$

I find no other way but to accept this by definition even though there's like $\displaystyle \infty-\infty$ involved.

I'm afraid situation here might be relatively similar :(

That integral does not exist, either. If you want to mess around with Cauchy Principal Values, you might get some meaningful answer, probably zero as you've stated. However, the function simply is not Riemann integrable.
• Mar 10th 2011, 10:30 AM
topsquark
I'm inclined to go along with along with WolframAlpha. $\displaystyle \displaystyle \lim_{\gamma \to \infty} \left ( ln^2(\gamma) - ln^2(\gamma) \right ) = \lim_{\gamma \to \infty} 0 = 0$

-Dan
• Mar 10th 2011, 10:34 AM
Ackbeet
Quote:

Originally Posted by topsquark
I'm inclined to go along with along with WolframAlpha. $\displaystyle \displaystyle \lim_{\gamma \to \infty} \left ( ln^2(\gamma) - ln^2(\gamma) \right ) = \lim_{\gamma \to \infty} 0 = 0$

-Dan

I would claim that's a Cauchy Principal Value of the integral. Normally, if I'm not mistaken, you have to evaluate the limits first, and then subtract.