# Integral $\int_0^{\infty } \frac{1}{(\alpha x^2 + 1) \left(- 2 \sqrt{\frac{ x^2}{x^2+1}}+2 x+\pi \right)} \, dx$

Does the following integral admit a closed-form expression?

$$\int_0^{\infty } \frac{1}{(\alpha x^2 + 1) \left(- 2 \sqrt{\frac{ x^2}{x^2+1}}+2 x+\pi \right)} \, dx \;\; , \;\; 0 \leq \alpha \leq 1.$$

Mathematica and Maple couldn't derive solutions in terms of elementary functions. The numerical value with $\alpha = 1$ is approx. 0.361.

• Substitute $x=\frac{2y}{1-y^2}$ to transform the integral to $2\int^1_0\frac{(1-y^2)dy}{8y^3+\pi(1-y^4)}$. The closed form involves the root of the polynomial $8y^3+\pi(1-y^4)$. Mar 10, 2014 at 22:41
• My apologies, the integral was missing the constant $\alpha$ - is the modified integral tractable?
– armn
Mar 11, 2014 at 9:01
• Even less tractable. Mar 11, 2014 at 12:09

We need to evaluate the integral. It's more convenient to use $a=1/ \alpha$:

$$I(a)=a \int_0^\infty \frac{dx}{(x^2+a)(\pi+2x-2x/\sqrt{1+x^2})}$$

We can make a substitution:

$$x=\sinh t$$

Then:

$$I(a)=a \int_0^\infty \frac{\cosh^2 t dt}{(\sinh^2 t+a)(\pi \cosh t+\sinh 2t-2\sinh t)}$$

Now we express the hyperbolic functions in their exponential form and make a substitution:

$$p=e^{-t}$$

Then we obtain a rational integral:

$$I(a)=2a \int_0^1 \frac{p (1+p^2)^2 dp}{(1+2(2a-1) p^2+p^4) (1+(\pi-2)p+(\pi+2) p^3-p^4)}$$

A special value $a=1$ which the OP used as an example, gives a more simple integral:

$$I(1)=2 \int_0^1 \frac{p dp}{(1+(\pi-2)p+(\pi+2) p^3-p^4)}$$

The general case can be evaluated explicitly by factoring the polynomials in the denominator and using partial fractions. The result is going to be ugly.