improper integral

**Random Variable** · January 23rd 2011, 10:38 AM

Show that $\displaystyle \int^{\infty}_{0} \frac{\sin x}{x} \ \ln x \ dx = -\gamma \int_{0}^{\infty} \frac{\sin x}{x} \ dx = - \frac{\pi}{2} \ \gamma$ (where $\gamma$ is Euler's constant).

**Random Variable** · January 24th 2011, 07:01 AM

Hint:

Spoiler:

Bigger Hint:

Spoiler:

**Drexel28** · January 24th 2011, 10:26 PM

Originally Posted by Random Variable

Show that $\displaystyle \int^{\infty}_{0} \frac{\sin x}{x} \ \ln x \ dx = -\gamma \int_{0}^{\infty} \frac{\sin x}{x} \ dx = - \frac{\pi}{2} \ \gamma$ (where $\gamma$ is Euler's constant).

I would proceed as follows:

Spoiler:

**Random Variable** · January 24th 2011, 10:56 PM

It's amazing how I don't realize that many of the problems I post are interconnected until you point it out to me.

$\displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^2} \ dx = 0$

$\displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^{2}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} \ln x \ \mathcal{L} \{ \sin t \} (x) \ dt \ dx = 0$

$\displaystyle \int^{\infty}_{0} \sin t \int^{\infty}_{0} e^{-xt} \ln x \ dx \ dt =0$

$\displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \Big(\int^{\infty}_{0} e^{-u} \ln u \ du - \ln t \int^{\infty}_{0} e^{-u}\ du\Big) \ dt =0$

$\displaystyle \int^{\infty}_{0} \frac{\sin t}{t} (-\gamma - \ln t) dt = 0$

so $\displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \ln t \ dt = -\gamma \int^{\infty}_{0} \ \frac{\sin t}{t} \ dt$

**Drexel28** · January 24th 2011, 10:58 PM

Originally Posted by Random Variable

It's amazing how I don't realize that many of the problems I post are interconnected until you point it out to me.

$\displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^2} \ dx = 0$

$\displaystyle \int^{\infty}_{0} \frac{\ln x}{1+x^{2}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} \ln x \ \mathcal{L} \{ \sin t \} (x) \ dt \ dx = 0$

$\displaystyle \int^{\infty}_{0} \sin t \int^{\infty}_{0} e^{-xt} \ln x \ dx \ dt =0$

$\displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \Big(\int^{\infty}_{0} e^{-u} \ln u \ du - \ln t \int^{\infty}_{0} e^{-u}\ du\Big) \ dt =0$

$\displaystyle \int^{\infty}_{0} \frac{\sin t}{t} (-\gamma - \ln t) dt = 0$

so $\displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \ln t \ dt = -\gamma \int^{\infty}_{0} \ \frac{\sin t}{t} \ dt$

Yeah, but I'm often lazy and just look for the quickest most mechanical solution, you take time and come up with neat non-obvious solutions. So, in the end, which is better?

**Random Variable** · January 26th 2011, 08:38 PM

Can we justify changing the order of integration, or is my solution invalid?

**uniquesailor** · February 16th 2011, 12:58 PM

**TheCoffeeMachine** · February 18th 2011, 04:49 AM

Originally Posted by uniquesailor

...

What is 'MAZ identity', if you don't mind me asking?

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