I tried with contour integral:

$$I = \frac{1}{2}\int_{-\infty}^{\infty } dz z^2\left(\dfrac{z^2 +1}{\sqrt{z^4 + 2 z^2 }} - 1\right)$$

The contour can be deformed into the upper half plane. But there is a cut from $\sqrt{2} i $ up to $\infty i $. The integral along the cut is no simpler than the original integral.

How to proceed? The answer is $\sqrt{2}/3$.

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    $\begingroup$ Please write it in a clearer way, for it's not clear if the $-1$ term is inside or outside the denominator. $\endgroup$ – Turing Jan 16 at 13:37
  • $\begingroup$ Ok, after having calculated it, the minus one is out, otherwise it does not converge. Going to edit for the sake of clearness. $\endgroup$ – Turing Jan 16 at 13:55

Let $$I_n=\int_0^n \frac{y(y^2+1)}{\sqrt{y^2+2}}dy-\int_0^ny^2 dy $$

In the first integral, substitute $y^2+2 =t^2 \implies 2y dy =2t dt $.

$$I_n= \int_{\sqrt 2}^{\sqrt{n^2+2}}( t^2-1) dt -\int_0^n y^2 dy \\ =\frac{(n^2+2)^{3/2} -2\sqrt 2}{3}+\sqrt 2-\sqrt{n^2+2}-\frac{n^3}{3}$$

Since your integrand is even, $$I=\lim_{n\to \infty} I_n =\frac{\sqrt 2}{3} +\frac 13 \lim_{n\to\infty} (n^2+2)^{3/2} -n^3 -3\sqrt{n^2+2}\\ $$ All it remains now is to show that the latter limit is $0$. It can be written as $$(n^2-1)\sqrt{n^2+2} -n^3 \\ = (n^3-n) \sqrt{1+\frac{2}{n^2}} -n^3 \\ \to (n^3-n)\bigg( 1+\frac{1}{n^2} +\text{smaller terms}\bigg) -n^3 \\ =n^3-n+n-n^3 +\frac 1n \to 0$$

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    $\begingroup$ I think a complex-analytical solution is desired here. $\endgroup$ – Parcly Taxel Jan 16 at 14:04
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    $\begingroup$ nice solution! This kind of approach is completely new to me. $\endgroup$ – S. Kohn Jan 16 at 23:16
  • $\begingroup$ @ParclyTaxel I don’t think so. $\endgroup$ – Tavish Jan 17 at 8:39

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