# improper integral

• Jan 23rd 2011, 10:38 AM
Random Variable
improper integral
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).
• Jan 24th 2011, 07:01 AM
Random Variable
Hint:

Spoiler:
Start with the fact that $\displaystyle \int^{\infty}_{0} \frac{ \ln x} {1+x^{2}} \ dx = 0$

Bigger Hint:

Spoiler:
$\displaystyle \int^{\infty}_{0} e^{-x} \ln x = - \gamma$
• Jan 24th 2011, 10:26 PM
Drexel28
Quote:

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:

Let $\displaystyle I(\alpha)=\int_0^{\infty}\sin(x)x^{\alpha-1}\text{ }dx$. By previous problem we have that $\displaystyle I(\alpha)=\frac{\pi}{2\Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)}$ for $\alpha\in(-1,1)$ and thus on that interval

\displaystyle \begin{aligned}I'(\alpha) &=\frac{-\pi\left[-2\Gamma'(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)-\pi\Gamma(1-\alpha)\sin\left(\frac{\pi\alpha}{2}\right)\right]}{\left(2\Gamma\left(1-\alpha\right)\cos\left(\frac{\pi\alpha}{2}\right)\ right)^2}\\ &=\frac{\pi}{4}\frac{2\Gamma'(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)+\pi\Ga mma(1-\alpha)\sin\left(\frac{\pi\alpha}{2}\right)}{\left (\Gamma(1-\alpha)\cos\left(\frac{\pi\alpha}{2}\right)\right) ^2}\end{aligned}

That said, using Lebniz's differentiation under the integral technique (checking that it's applicable) we find that

$\displaystyle I'(\alpha)=\int_0^{\infty}\frac{\partial}{\partial \alpha}\sin(x)x^{\alpha-1}\text{ }dx=\int_0^\infty} \sin(x)x^{\alpha-1}\log(x)\text{ }dx$

Thus, our integral is $\displaystyle I'(0)=\frac{\pi}{4}\cdot2\Gamma'(1)=\frac{-\pi\gamma}{2}$.

• Jan 24th 2011, 10:56 PM
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 $\displaystyle \int^{\infty}_{0} \frac{\sin t}{t} \ln t \ dt = -\gamma \int^{\infty}_{0} \ \frac{\sin t}{t} \ dt$
• Jan 24th 2011, 10:58 PM
Drexel28
Quote:

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?
• Jan 26th 2011, 08:38 PM
Random Variable
Can we justify changing the order of integration, or is my solution invalid?
• Feb 16th 2011, 12:58 PM
uniquesailor
Solution By use of MAZ Identity
• Feb 18th 2011, 04:49 AM
TheCoffeeMachine
Quote:

Originally Posted by uniquesailor
...

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