Seeking methods to find $\int _0^{\infty }\frac{x^{\beta }}{x^2\left(x-t\right)^2}\:dx$ I was able to prove with real methods that:
$$\int _0^{\infty }\frac{x^{\beta }}{x^2\left(x-t\right)^2}\:dx=\left(\beta -2\right)\left(-t\right)^{\beta -3}\:\pi \csc \left(\pi \beta \right)$$
Which only converges when $t<0$ and $1<\beta<3$.
Im really interested in seeing other kind of approaches so please share them.
My Attempt.
I first considered the following integral,
$$\int _0^{\infty }\frac{x^a}{x^b+c}\:dx\overset{x=xc^{\frac{1}{b}}}=c^{\frac{a+1}{b}-1}\int _0^{\infty }\frac{x^a}{x^b+1}\:dx$$
$$=c^{\frac{a+1}{b}-1}\:\frac{\pi }{b}\csc \left(\pi \frac{a+1}{b}\right)$$
Now let's call the integral $I\left(c\right)$ and differentiate.
$$I\left(c\right)=\int _0^{\infty }\frac{x^a}{x^b+c}\:dx=c^{\frac{a+1}{b}-1}\:\frac{\pi }{b}\csc \left(\pi \frac{a+1}{b}\right)$$
$$I'\left(c\right)=\int _0^{\infty }\frac{x^a}{\left(x^b+c\right)^2}\:dx=-\left(\frac{a+1}{b}-1\right)c^{\frac{a+1}{b}-2}\:\frac{\pi }{b}\csc \left(\pi \frac{a+1}{b}\right)$$
Using this result we can now find the original integral by just plugging the values for $a$,$b$ and $c$:
$$\int _0^{\infty }\frac{x^{\beta -2}}{\left(x-t\right)^2}\:dx=-\left(\beta -2+1-1\right)\left(-t\right)^{\beta -2+1-2}\:\pi \csc \left(\pi \left(\beta -2+1\right)\right)$$
$$=\left(\beta -2\right)\left(-t\right)^{\beta -3}\:\pi \csc \left(\pi \beta \right)$$
So,
$$\boxed{\int _0^{\infty }\frac{x^{\beta }}{x^2\left(x-t\right)^2}\:dx=\left(\beta -2\right)\left(-t\right)^{\beta -3}\:\pi \csc \left(\pi \beta \right)}$$
 A: For a residue approach, let $s=\beta-2$ and
$$\mathcal I=\int_0^\infty\frac{x^s}{(x-t)^2}~\mathrm dx$$
$$\mathcal J=\int_0^\infty\frac{x^s\ln(x)}{(x-t)^2}~\mathrm dx$$
Apply a keyhole contour to $\mathcal J$ to get
$$\mathcal J-e^{2\pi is}(\mathcal J+2\pi i\mathcal I)=2\pi i\mathop{\rm Res}\limits_{x=t}\frac{x^s\ln(x)}{(x-t)^2}=2\pi i\lim_{x\to t}\frac{\mathrm d}{\mathrm dx}x^s\ln(x)=2\pi it^{s-1}(s\ln(t)+1)$$
Let $c=-t$ to simplify the RHS to
$$\mathcal J-e^{2\pi is}(\mathcal J+2\pi i\mathcal I)=-2\pi ie^{\pi is}c^{s-1}(s\ln(c)+1+\pi is)$$
All that remains is to write out the real and imaginary parts and solve for $\mathcal I$.
A: This can be nicely done using Ramanujan's master theorem, which states (source is Wikipedia)

If a complex-valued function $ f(x) $ has an expansion of 
   the form
   $$ f(x)=\sum_{k=0}^\infty \frac{\varphi(k)}{k!}(-x)^k $$
   then the Mellin transform of $f(x)$ is given by
   $$ \int_0^\infty x^{s-1} f(x) \, dx = \Gamma(s)\varphi(-s) $$
   where $ \Gamma(s) $ is the gamma function.

Your integral is
$$I=\int_0^\infty x^{s-1} f(x) dx $$
with $s=\beta-1$ and $f(x)=(x-t)^{-2}$. Computing derivatives of $f(x)$ at $x=0$, you can easily establish its Taylor series expansion
$$\begin{align*}f(x)&=(x-t)^{-2}=\sum_{k=0}^\infty x^k (1+k) t^{-2-k}\\
&=\sum_{k=0}^\infty \frac{(-x)^k}{k!} \underbrace{\Gamma(1+k) (-1)^k (1+k) t^{-2-k}}_{=:\varphi(k)}
\end{align*}$$
In the last equality we used that $\Gamma(1+k)=k!$. Now Ramanujan's master theorem tells us that
$$
\begin{align*}
I&=\int_0^\infty x^{s-1} f(x) dx \\ 
& = \Gamma(s) \varphi(-s) \\
&=\Gamma(s) \Gamma(1-s) (-1)^s (1-s) t^{-2+s} \\
&= \frac{\pi}{\sin(\pi s)}(-1)^s (1-s) t^{s-2} 
\end{align*}
$$
In the last step, we applied Euler's reflection formula for the Gamma function 
 $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}$ (which you can also find on Wikipedia).
Finally, we can let $s=\beta-1$ to find
$$ I = (-1)^\beta (2-\beta) t^{\beta-3} \frac{\pi}{\sin(\pi \beta)} $$
where we used $\sin(\pi(\beta-1))=-\sin(\pi \beta)$.
A: Well, we are trying to solve the following integral:
$$\mathcal{I}_\beta\left(\gamma\right):=\int_0^\infty\frac{x^\beta}{x^2\left(x-\gamma\right)^2}\space\text{d}x\tag1$$
Using the evaluating integrals over the positive real axis property of the Laplace transform we can write:
$$\mathcal{I}_\beta\left(\gamma\right)=\int_0^\infty\mathcal{L}_x\left[x^\beta\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{x^2\left(x-\gamma\right)^2}\right]_{\left(\text{s}\right)}\space\text{ds}\tag2$$
Now, using the convolution property of the Laplace transform we can write:
$$\mathcal{L}_x^{-1}\left[\frac{1}{x^2\left(x-\gamma\right)^2}\right]_{\left(\text{s}\right)}=\int_0^\text{s}\mathcal{L}_x^{-1}\left[\frac{1}{x^2}\right]_{\left(\tau\right)}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{\left(x-\gamma\right)^2}\right]_{\left(\text{s}-\tau\right)}\space\text{d}\tau\tag3$$
Using the table of selected Laplace transforms, we can write:


*

*$$\mathcal{L}_x^{-1}\left[\frac{1}{x^2}\right]_{\left(\tau\right)}=\tau\tag4$$

*$$\mathcal{L}_x^{-1}\left[\frac{1}{\left(x-\gamma\right)^2}\right]_{\left(\text{s}-\tau\right)}=\left(\text{s}-\tau\right)\exp\left(\gamma\left(\text{s}-\tau\right)\right)\tag5$$
So:
$$\int_0^\text{s}\mathcal{L}_x^{-1}\left[\frac{1}{x^2}\right]_{\left(\tau\right)}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{\left(x-\gamma\right)^2}\right]_{\left(\text{s}-\tau\right)}\space\text{d}\tau=\int_0^\text{s}\tau\left(\text{s}-\tau\right)\exp\left(\gamma\left(\text{s}-\tau\right)\right)\space\text{d}\tau=$$
$$\frac{2+\text{s}\gamma+\exp\left(\text{s}\gamma\right)\left(\text{s}\gamma-2\right)}{\gamma^3}\tag6$$
So, we get:
$$\mathcal{I}_\beta\left(\gamma\right)=\int_0^\infty\frac{\Gamma\left(1+\beta\right)}{\text{s}^{1+\beta}}\cdot\frac{2+\text{s}\gamma+\exp\left(\text{s}\gamma\right)\left(\text{s}\gamma-2\right)}{\gamma^3}\space\text{ds}=$$
$$\frac{\Gamma\left(1+\beta\right)}{\gamma^3}\int_0^\infty\frac{2+\text{s}\gamma+\exp\left(\text{s}\gamma\right)\left(\text{s}\gamma-2\right)}{\text{s}^{1+\beta}}\space\text{ds}\tag7$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\left.\int_{0}^{\infty}{x^{\beta} \over x^{2}\pars{x - t}^{2}}\,\dd x
\,\right\vert_{\ {\Large t\ <\ 0} \atop
{\large\vphantom{A^{A}} 1\ <\ \beta\ <\ 3}}}
\\[5mm] \stackrel{x/\verts{t}\ \mapsto\ x}{=}\,\,\,&
\verts{t}^{\,\beta - 3}
\int_{0}^{\infty}{x^{\beta - 2} \over \pars{x + 1}^{2}}\,\dd x
\\[5mm] = &\
\verts{t}^{\,\beta - 3}
\int_{1}^{\infty}{\pars{x - 1}^{\beta - 2} \over x^{2}}\,\dd x
\\[5mm] 
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,&
\verts{t}^{\,\beta - 3}\int_{1}^{0}{\pars{1/x - 1}^{\beta - 2} \over \pars{1/x}^{2}}\,
\pars{-\,{\dd x \over x^{2}}}
\\[5mm] = &\
\verts{t}^{\,\beta - 3}\int_{0}^{1}x^{2 - \beta}\pars{1 - x}^{\beta - 2}\,\dd x
\\[5mm] = &\
\verts{t}^{\,\beta - 3}\,{\Gamma\pars{3 - \beta}\Gamma\pars{\beta - 1} \over \Gamma\pars{2}}
\\[5mm] = &\
\verts{t}^{\,\beta - 3}\pars{2- \beta}
\bracks{\vphantom{\Large A}\Gamma\pars{2 - \beta}\Gamma\pars{\beta - 1}}
\\[5mm] = &\
\verts{t}^{\,\beta - 3}\pars{2- \beta}\,
{\pi \over \sin\pars{\pi\bracks{2 - \beta}}}
\\[5mm] = &\
\bbx{\pars{\beta - 2}
\pars{-t}^{\,\beta - 3}\,\pi\csc\pars{\pi\beta}} \\ &
\end{align}
