<s>[INST] 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
 [/INST] 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
$$\int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \binom{2(n-1)}{n-1}\frac{\pi}{4^{n-1}(a+1)^{n - \frac{1}{2}}}$$
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