Evaluate $\int_{-1}^{1}\mathop{dx}\frac{1}{(x^2+a^2)^k}$?

Question

My original question was what is the condition for the following integral to converge: $$\int_{-1}^{1}\mathop{dx}\left[\frac{1}{x^{2k}}-\frac{1}{(x^2+a^2)^k}\right].$$ I know $k>0$. By motivation of Dr. MV, the condition is simply $2k<1$. Now, my question is how do I evaluate the second part of the above integral, that is $$\int_{-1}^{1}\mathop{dx}\frac{1}{(x^2+a^2)^k},$$ for any $k<1/2$ and real $a$. Any idea?

Answer 1

In general, for $a,k\in\mathbb R^\star$, we have

$$I~=~\dfrac2{a^{2k}}\cdot~_2F_1\bigg(\dfrac12~,~k~;~\dfrac32~;~-\dfrac1{a^2}\bigg)~=~-\dfrac i{|a|^{2k-1}}\cdot B\bigg(-\dfrac1{a^2}~,~\dfrac12~,~1-k\bigg),$$

which, needless to say, does not say much. For starters, let $x=at$, and then use the parity of the integrand. Afterwards, in order to obtain the first expression, expand the integrand into its own binomial series, and reverse the order of summation and integration. As for the latter, a second $($trigonometric$)$ substitution of the form $t=\tan u$ is in order, to reduce the expression to a Wallis integral, whose relation to the $($ incomplete $)$ beta function is well-known. For $k\in\mathbb N^\star$ and $a=1$, we have

$$I~=~\dfrac{A_{k-1}}{(2k-2)!!}~+~\dfrac1{2^{2k-2}}{2k-2\choose k-1}\dfrac\pi2,$$

where $A_{k-1}$ form this sequence.