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Going through some notes I seem to have used the following expression:

$$\tag{1} \int_0^{\infty}u^k(1+u)^{n-k}(1+2u)^{-n-3}du=B(k+1, 2)\ _2F_1(n+3, k+1;k+3;-1)$$

where $k, n\in\{2,4,6,\ldots\}$, $B(a,b)$ is the Beta function and $_2F_1(a,b;c;z)$ is the hypergeometric function.

The expression (1) seems to be true when checking individual values, but I can't figure out where I got it. It is similar to the following two from here:

$$\tag{2} \int_0^{1}u^{k-1}(1-u)^{n-k-1}(1-zu)^{-m}du=\frac{\Gamma(n)}{\Gamma(k)\Gamma(n-k)}\ _2F_1(m, k; n; z)$$

$$\tag{3} \int_0^{\infty}u^{n-k-1}(1+u)^{m-n}(1+u-z)^{-m}du=\frac{\Gamma(n)}{\Gamma(k)\Gamma(n-k)}\ _2F_1(m, k; n; z)$$

However, I haven't been able to deduce (1) from (2) or (3). This is a problem because I now need to calculate a slightly different integral where I could use the general version of (1):

$$\tag{4} \int_0^{\infty}u^k(1+u)^{-1}(1+2u)^{-n-3}du$$

Anyone know of a general formula that reduces to (1)?

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  • 2
    $\begingroup$ The standard trick is use the substitution $x = \frac{u}{1+u}$ to turn the integral over $[0,\infty)$ to one over $[0,1]$. $\endgroup$ – achille hui Nov 6 '13 at 17:37
  • $\begingroup$ Thanks! That works perfectly $\endgroup$ – jorgen Nov 6 '13 at 17:42
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In case someone else ends up here: I found the answer in this book (page 60, equation (12)):

$\displaystyle\int_0^{\infty}u^{b-1}(1+u)^{c-b-1}(1+zu)^{-a}du=\frac{\Gamma(b)\Gamma(a+1-c)}{\Gamma(a+b+1-c)}\ _2F_1(a,b;a+b+1-c,1-z)$

EDIT: As pointed out by achille hui, this can be obtained from (2) in the OP by substituting $x=\frac{u}{1+u}$

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