$$\int_{0}^{\infty}{2\sqrt{x^2+1}+\gamma(x^2-x)\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{\mathrm dx\over (x^2+1)^2}=1\tag1$$

Where $\gamma=0.577...$ it is Euler's Constant

$$\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}+\gamma\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx\tag2$$

$$\gamma\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx=\gamma\int_{0}^{\infty}{x^2\over (x^2+1)^{5/2}}\mathrm dx-\gamma\int_{0}^{\infty}{x\over (x^2+1)^{5/2}}\mathrm dx\tag3$$

Enforcing $u=x^2$ then $du=2xdx$

Recalling from the beta function

$$\int_{0}^{\infty}{u^m\over (u+1)^{m+n+2}}du=B(m+1,n+1)\tag4$$

$${\gamma\over 2}\int_{0}^{\infty}{u\over (u+1)^{5/2}}\mathrm du-{\gamma\over 2}\int_{0}^{\infty}{1\over (u+1)^{5/2}}\mathrm du={B(3/2,1)\over2}-{B(1,3/2)\over2}=0\tag5$$

So this part of integration must be 1! It is ready done my @Marco on my previous post.

$$\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}\tag6$$

Can anyone prove this integral $(1)$ via another method? Thank you! Sorry for not addressing the question properly.

  • $\begingroup$ Euler's constant is usually $e$. The constant you have is usually called the Euler-mascheroni constant. $\endgroup$ – Simply Beautiful Art Jan 9 '17 at 16:17
  • $\begingroup$ This looks like very similar to a previous post: math.stackexchange.com/questions/2090206/… $\endgroup$ – Math-fun Jan 9 '17 at 16:17
  • 1
    $\begingroup$ What is your question, really? Could you clarify? $\endgroup$ – mickep Jan 9 '17 at 16:18
  • $\begingroup$ @SimpleArt See this $\endgroup$ – Mark Viola Jan 9 '17 at 16:20
  • $\begingroup$ @Simple Art: I have seen $\gamma$ called "Euler's constant" about ten times more frequently than "the Eueler-Mascheroni constant." Perhaps I am biased by work in mathematical physics, where $\gamma$ plays a crucial role in the $\overline{MS}$ renormalization prescription. Also, $e$ is rarely referred to as "Euler's constant" since it is so familiar, most literature just uses "$e$". $\endgroup$ – Mark Fischler Jan 9 '17 at 16:24

In fact, letting $x=\tan t$ and $u=\sin t$, one has \begin{eqnarray} &&\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}\\ &=&2\int_0^{\pi/2}\frac{1}{\sqrt{1+\sec^2t}}\frac{1}{\sec^4t}\sec^2tdt\\ &=&2\int_0^{\pi/2}\frac{\cos^3 t}{\sqrt{1+\sec^2t}}dt\\ &=&2\int_0^{\pi/2}\frac{\cos^3 t}{\sqrt{2-\sin^2t}}dt\\ &=&2\int_0^{1}\frac{1-u^2}{\sqrt{2-u^2}}du\\ &=&u\sqrt{2-u^2}|_0^1\\ &=&1.\\ &&\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx\\ &=&\int_{0}^{\pi/2}{\tan^2t-\tan t\over (\sec^2t)^{5/2}}\sec^2t\mathrm dx\\ &=&\int_{0}^{\pi/2}(\tan^2t-\tan t)\cos t\mathrm dx\\ &=&\int_{0}^{\pi/2}\cos t\sin t(\sin t-\cos t)\mathrm dx\\ &=&\int_{0}^{\pi/2}\cos t\sin^2 t\mathrm dx-\int_{0}^{\pi/2}\cos^2 t\sin t\mathrm dx\\ &=&0. \end{eqnarray}


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