Help Finding Integral: x^2 + b^2 to the nth Power

[Wired55]

Need help finding

$$\int$$1/ (x$$^{}2^{}$$ + b$$^{}2^{}$$ )$$^{}n^{}$$ dx

with limits negative infinity to infinity

where b, n some constant

No work is required since its just part of a quantum mechanics problem, i can't find the integral in any tables and i don't have mathematica or anything available to me right now.

edit: or if anyone can link me to a page with definite integrals of that form, as i also may need the same thing multiplied by x^n

thanks

 

[Kreizhn]

Unless I'm mistaken, the problem will depend on whether or not n is even or odd.

let $$x= b \tan{\theta}$$

then $$(x^2 + b^2)^n = b^{2n}(\tan^2{\theta} + 1)^n$$
$$dx = b \sec^2 {\theta}$$

thus

$$\displaystyle \int \frac{1}{(x^2+b^2)^n} dx = \frac{1}{b^{2n-1}}\int \frac{d\theta}{\sec^{n-2}{\theta}} d\theta$$

$$= \frac{1}{b^{2n-1}} \int cos^{n-2}{\theta} d\theta$$

Now for simplicity sake, let $$k= n-2$$

If k is even (iff n is even) then use the identity

$$\cos^2 {x}= \frac{1}{2} ( 1 + \cos{2x})$$

If k is odd (iff n is odd) then take

$$\cos^k{x} = \cos{x}(1-\sin^2{x})^{k-1}$$ and use basic substitution.

Note that (k-1) is even since k is odd.