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I ran across an interesting integral and I am wondering if anyone knows where I may find its derivation or proof. I looked through the site. If it is here and I overlooked it, I am sorry.

$$\displaystyle\frac{1}{\int_{-\infty}^{\infty}\frac{1}{(e^{x}-x)^{2}+{\pi}^{2}}dx}-1=W(1)=\Omega$$

$W(1)=\Omega$ is often referred to as the Omega Constant. Which is the solution to

$xe^{x}=1$. Which is $x\approx .567$

Thanks much.

EDIT: Sorry, I had the integral written incorrectly. Thanks for the catch.

I had also seen this:

$\displaystyle\int_{-\infty}^{\infty}\frac{dx}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+W(1)}=\frac{1}{1+\Omega}\approx .638$

EDIT: I do not what is wrong, but I am trying to respond, but can not. All the buttons are unresponsive but this one. I have been trying to leave a greenie and add a comment, but neither will respond. I just wanted you to know this before you thought I was an ingrate. Thank you. That is an interesting site.

  • 1
    Maple says the integral is approximately 0.605, but $1/(1+W(1))$ is approximately 0.6382011-06-16
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    In Maple I get .6381037434, which agrees with 1/(1+LambertW(1)). I guess the 0.605 refers to a version of the integral before the last edit.2011-06-16
  • 0
    Originally it had $+x$ instead of $-x$.2011-06-16
  • 0
    Nice question (+1)2012-10-09

4 Answers 4