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

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March 10th 2011, 10:00 AM
Pranas

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

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

$\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?
March 10th 2011, 10:06 AM
Ackbeet

Couple of things:

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

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

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

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

Quote:

Originally Posted by Ackbeet

Couple of things:

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

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

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

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

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

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

Quote:

Originally Posted by Pranas

Somewhat of a similar behaviour:

$\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 $\infty-\infty$ 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.
March 10th 2011, 10:30 AM
topsquark

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

-Dan
March 10th 2011, 10:34 AM
Ackbeet

Quote:

Originally Posted by topsquark

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

-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.