How to integrate the following definite integral?

Question

$$\int_0^\infty\frac{(B+W)^{-k}}{\sqrt{W+\varphi}} \, dW$$ Is there any general result available . I referred to some table of integrals. There I didn't find a direct result. But found that there are results available if I split this integral into two, ie from 0 to $\varphi$ and then $\varphi$ to $\infty$. Now when I use those results I am getting a solution which does not matches with the results that I obtained using mathematica... Please help me out

Answer 1

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Define $I_{n}{\left(a,b\right)}$ via the integral representation

$$I_{n}{\left(a,b\right)}:=\int_{0}^{\infty}\frac{\left(a+x\right)^{-n}}{\sqrt{b+x}}\,\mathrm{d}x;~~~\small{n>\frac12\land a>0\land b>0}.$$

Letting $n>\frac12\land a>0\land b>0$, we can readily evaluate the integral in question in terms of the Gauss hypergeometric function*:

$$\begin{align} I_{n}{\left(a,b\right)} &=\int_{0}^{\infty}\frac{\left(a+x\right)^{-n}}{\sqrt{b+x}}\,\mathrm{d}x\\ &=\int_{0}^{\infty}\frac{\sqrt{b}}{\left(a+by\right)^{n}\sqrt{1+y}}\,\mathrm{d}y;~~~\small{\left[x=by\right]}\\ &=\sqrt{b}\int_{0}^{1}\frac{t^{n-2}\sqrt{t}}{\left(b+at-bt\right)^{n}}\,\mathrm{d}t;~~~\small{\left[y=\frac{1-t}{t}\right]}\\ &=\frac{\sqrt{b}}{b^{n}}\int_{0}^{1}\frac{t^{n-\frac32}}{\left[1-\left(\frac{b-a}{b}\right)t\right]^{n}}\,\mathrm{d}t\\ &=\frac{2\sqrt{b}}{\left(2n-1\right)b^{n}}\,{_2F_1}{\left(n,n-\frac12;n+\frac12;\frac{b-a}{b}\right)}.\blacksquare\\ \end{align}$$

The above result holds for any real $n$ greater than $\frac12$. If we are free to assume that $n$ is a positive integer, the hypergeometric function will simplify to a sum of elementary functions as we might expect.

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*Note: the Gauss hypergeometric function can be defined via the integral representation

$${_2F_1}{\left(a,b;c;z\right)}:=\frac{1}{\operatorname{B}{\left(b,c-b\right)}}\int_{0}^{1}\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}}\,\mathrm{d}t,$$

where $0<\Re{\left(b\right)}<\Re{\left(c\right)}\land z\in\left(-\infty,1\right)$.