Math Help - improper integral #2

1. 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.

2. 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!

3. 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.

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

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

4. 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.

5. 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