# $I = \int_0^\infty y^2 [ (y^2 +1)/\sqrt{y^4 + 2 y^2 } - 1] dy$

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 $$\sqrt{2}/3$$.

• Please write it in a clearer way, for it's not clear if the $-1$ term is inside or outside the denominator. – Turing Jan 16 at 13:37
• Ok, after having calculated it, the minus one is out, otherwise it does not converge. Going to edit for the sake of clearness. – 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 $$(n^2-1)\sqrt{n^2+2} -n^3 \\ = (n^3-n) \sqrt{1+\frac{2}{n^2}} -n^3 \\ \to (n^3-n)\bigg( 1+\frac{1}{n^2} +\text{smaller terms}\bigg) -n^3 \\ =n^3-n+n-n^3 +\frac 1n \to 0$$