How can i solve this integral by differentiating under integral sign or by any other method $$\int_{1}^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}}$$
where $\Im(a)\neq0$ and $\Re\ge-1$
Here's how i tried to solve
$$I(a)=\int_{1}^{\infty}\frac{dx}{(x+a)\sqrt{x-1}}=\frac{\pi}{\sqrt{a+1}}$$
Differentiating wrt $a$
$$I'(a)=\int_{1}^{\infty}\frac{dx}{(x+a)^2\sqrt{x-1}}=\frac{\pi}{2(a+1)^{3/2}}$$
$$I''(a)=\int_{1}^{\infty}\frac{dx}{(x+a)^3\sqrt{x-1}}=\frac{3\pi}{8(a+1)^{5/2}}$$
$$I'''(a)=\int_{1}^{\infty}\frac{dx}{(x+a)^4\sqrt{x-1}}=\frac{5\pi}{16(a+1)^{7/2}}$$
By continuing in similar way how can i obtain the general expression for it's $n$th derivative
 A: A closed solution can  be obtained for any $n>\frac{1}{2}$ (not necessarily positive integer).
$I(n,a)=\int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}}=\int_0^{\infty}\frac{ds}{(s+a+1)^n\sqrt{s}}=\frac{1}{(1+a)^{n-1/2}}\int_0^{\infty}\frac{dt}{(t+1)^n\sqrt t}$
Making change $x=\frac{1}{1+t}$
$$I(n,a)=\frac{1}{(1+a)^{n-1/2}}\int_0^1(1-x)^{-1/2}x^{n-3/2}dx=\frac{1}{(1+a)^{n-1/2}}B\Bigl(\frac{1}{2};n-\frac{1}{2}\Bigr)$$
$$=\frac{\sqrt \pi}{(1+a)^{n-1/2}}\frac{\Gamma\Bigl(n-\frac{1}{2}\Bigr)}{\Gamma(n)}$$
For $n\leqslant\frac{1}{2}$ integral diverges at $x\to \infty$ - this is clear from its initial form.
A: Upon differentiating, you should notice that
$$(n-1)! \int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \frac{(2n-3)\cdot(2n-5) \cdots1}{2\cdot2\cdots2}\frac{\pi}{(a+1)^{n-\frac{1}{2}}}$$
where there are $(n-1) \; 2's$ in product in the denominator.
Now the RHS can be manipulated a bit by multiplying and dividing by $(2n-2) \cdot (2n-4) \cdots 2 = 2^{n-1}\cdot (n-1)!$, which gives us
$$(n-1)! \int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \frac{(2n-2)!}{2^{n-1}\cdot2^{n-1}\cdot {(n-1)}!}\frac{\pi}{(a+1)^{n-\frac{1}{2}}}$$
which finally gives us
$$\int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \binom{2(n-1)}{n-1}\frac{\pi}{4^{n-1}(a+1)^{n - \frac{1}{2}}}$$
