# improper integral #2

• March 2nd 2011, 05:37 PM
Random Variable
improper integral #2
Show that $\displaystyle \int_{0}^{\infty} \Big( \frac{1}{1+x^{2}} - \cos x \Big) \ \frac{dx}{x}$ is an integral representation of Euler's constant.
• March 4th 2011, 10:33 PM
TheCoffeeMachine
Any hints, RV? The integrand can be written as:

$\displaystyle \sum_{k = 0}^{\infty}\frac{(-1)^k\left((2k)!-1\right)x^{2k-1}}{(2k)!}$

but that's probably just useless for this integral!
• March 5th 2011, 08:53 AM
Random Variable
Since my proof may not even be valid, I'll just post it instead of giving any hints.

$\displaystyle \int^{\infty}_{0} \Big( \frac{1}{1+x^{2}} - \cos x \Big) \ \frac{dx}{x}$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{1}{1+x^{2}} - \cos x \Big) e^{-tx} \ dt \ dx$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{1}{1+x^{2}} - \cos x \Big) e^{-tx} \ dx \ dt$ Probably justifiable.

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{e^{-tx}}{1+x^{2}} \ dx - \mathcal{L} \{\cos x \} (t) \Big) \ dt$

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{e^{-tx}}{1+x^{2}} \ dx - \frac{t}{1+t^{2}} \Big) \ dt$

let $u = tx$

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{t e^{-u}}{t^{2}+u^{2}} \ du - \frac{t}{1+t^{2}} \Big) \ dt$

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{t e^{-u}}{t^{2}+u^{2}} \ du - \int^{\infty}_{0} \frac{te^{-u}}{1+t^{2}} \ du \Big) \ dt$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{t}{t^{2}+u^{2}} - \frac{t}{1+t^{2}} \Big) e^{-u} \ du \ dt$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{t}{t^{2}+u^{2}} - \frac{t}{1+t^{2}} \Big) e^{-u} \ dt \ du$ Perhaps justifiable.

$\displaystyle = \frac{1}{2} \int^{\infty}_{0} \ln \Big(\frac{u^{2}+t^{2}}{1+t^{2}} \Big ) \Big|^{\infty}_{0} \ e^{-u} \ du$

$\displaystyle = - \int^{\infty}_{0} \ln u \ e^{-u} \ du = -(-\gamma) = \gamma$
• March 5th 2011, 09:06 AM
Random Variable
To justify the first change of order, you probably could write the integral as $\displaystyle \lim_{a \to 0} \int^{\infty}_{a} \int^{\infty}_{a} \Big( \frac{1}{1+x^{2}} - \cos x \Big) e^{-tx} \ dt \ dx \ , a>0$. Then change the order of integration, integrate with respect to x, and then justify bringing the limit inside of the integral.

For the second change, perhaps you can write it as $\displaystyle \lim_{b \to \infty} \int^{b}_{0} \int^{\infty}_{0} \Big( \frac{t}{u^{2}+t^{2}} - \frac{t}{1+ t^{2}} \Big) e^{-u} \ du \ dt$ where b is value such that the integrand is now always positive. Then do the same thing as above.

I think in both cases you could use use dominated convergence theorem to justify moving the integral inside of the integral. I would really like someone to tell me if I'm completely off base or not. This has been drving me a bit crazy over the past few days.
• March 5th 2011, 03:08 PM
uniquesailor
Quote:

Originally Posted by Random Variable
Since my proof may not even be valid, I'll just post it instead of giving any hints.

$\displaystyle \int^{\infty}_{0} \Big( \frac{1}{1+x^{2}} - \cos x \Big) \ \frac{dx}{x}$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{1}{1+x^{2}} - \cos x \Big) e^{-tx} \ dt \ dx$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{1}{1+x^{2}} - \cos x \Big) e^{-tx} \ dx \ dt$ Probably justifiable.

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{e^{-tx}}{1+x^{2}} \ dx - \mathcal{L} \{\cos x \} (t) \Big) \ dt$

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{e^{-tx}}{1+x^{2}} \ dx - \frac{t}{1+t^{2}} \Big) \ dt$

let $u = tx$

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{t e^{-u}}{t^{2}+u^{2}} \ du - \frac{t}{1+t^{2}} \Big) \ dt$

$\displaystyle = \int^{\infty}_{0} \Big( \int^{\infty}_{0} \frac{t e^{-u}}{t^{2}+u^{2}} \ du - \int^{\infty}_{0} \frac{te^{-u}}{1+t^{2}} \ du \Big) \ dt$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{t}{t^{2}+u^{2}} - \frac{t}{1+t^{2}} \Big) e^{-u} \ du \ dt$

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} \Big( \frac{t}{t^{2}+u^{2}} - \frac{t}{1+t^{2}} \Big) e^{-u} \ dt \ du$ Perhaps justifiable.

$\displaystyle = \frac{1}{2} \int^{\infty}_{0} \ln \Big(\frac{u^{2}+t^{2}}{1+t^{2}} \Big ) \Big|^{\infty}_{0} \ e^{-u} \ du$

$\displaystyle = - \int^{\infty}_{0} \ln u \ e^{-u} \ du = -(-\gamma) = \gamma$

Great (Clapping)