Integrating $t^{2r-1} / t^{2k} (1+t^2)^{r+1}$

Question

Let $k$ and $r$ be natural numbers such that $1 \leq k \leq r$. I want to calculate $$ \int_0^\infty \frac{t^{2r-1}}{t^{2k}(1+t^2)^{r+1}} dt. $$ Since the integrand is an odd function the standard residue theorem tricks I know don't work here. Wolfram Alpha also refused to calculate the definite integral and spits out an expression involving hypergeometric functions for the indefinite integral that I don't know how to handle. I'm sure this is very standard and obvious to people here and I'd appreciate any tips or hints.

In case anyone's interested this comes from trying to calculate the pushforward of the $(r+k)$-th power of the curvature form on the tautological line bundle on the projectivization of a given Hermitan holomorphic vector bundle of rank $r+1$ (hence the Fubini-Study volume factor $1/(1+t^2)^{r+1}$). The integral is the "trivial" part of the pushforward once converted into spherical coordinates (hence the $t^{2r-1}$).

Answer 1

Note that you need $k < r$ for the integral to converge. Then a sequence of standard substitutions brings it into a Beta function form:

$$\begin{align} \int_0^\infty \frac{t^{2(r-k)-1}}{(1+t^2)^{r+1}}\,dt &= \frac{1}{2} \int_0^\infty \frac{x^{r-k-1}}{(1+x)^{r+1}}\,dx\tag{$x = t^2$}\\ &= \frac{1}{2}\int_0^\infty \left(\frac{x}{1+x}\right)^{r-k-1}\left(\frac{1}{1+x}\right)^{k+2}\,dx \tag{$u = (1+x)^{-1}$}\\ &= \frac{1}{2}\int_0^1 u^k(1-u)^{r-k-1}\,du\\ &= \frac{1}{2} B(k+1,r-k)\\ &= \frac{\Gamma(k+1)\Gamma(r-k)}{2\Gamma(r+1)}\\ &= \frac{k!(r-k-1)!}{2\cdot r!}. \end{align}$$

Answer 2

Hint: Let $x=\dfrac1{1+t^2}$, then recognize the expression of the beta function in the new integral.

Answer 3

Using the following formulation of the Beta Function $$ \begin{align} \mathrm{B}(x,y)=\int_0^\infty\frac{r^{x-1}}{(1+r)^{x+y}}\,\mathrm{d}r\\ \end{align} $$ we get $$ \begin{align} \int_0^\infty\frac{t^{2r-1}}{t^{2k}(1+t^2)^{r+1}}\,\mathrm{d}t &=\frac12\int_0^\infty\frac{t^{2r-2}}{t^{2k}(1+t^2)^{r+1}}\,\mathrm{d}t^2\\ &=\frac12\int_0^\infty\frac{t^{r-k-1}}{(1+t)^{r+1}}\,\mathrm{d}t\\ &=\frac12\mathrm{B}(r-k,k+1)\\ &=\frac12\frac{\Gamma(r-k)\Gamma(k+1)}{\Gamma(r+1)} \end{align} $$ where we've employed the identity, proven in this answer, $$ \mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$

Answer 4

For the antiderivative, I found (I suppose) the same kind of result involving "nice" two $2F1$ hypergeometric functions but I was able to compute the integral. The result I obtained is $$ {\Gamma\left(k + 1\right)\Gamma\left(r - k\right) \over 2\Gamma\left(r + 1\right)} $$