How do we go from $\int_0^\infty{\frac{d \cos{2 \pi n u}}{(x-u)^4}}$? How do we go from $$-\frac{1}{2\pi n}\int_0^\infty{\frac{d \cos{2 \pi n u}}{(x-u)^4}}$$ to $$\frac{1}{2\pi n x^4}-\frac{1}{\pi^2 n^2}\int_0^\infty{\frac{d \sin{2 \pi n u}}{(x-u)^5}}$$
I found this in a paper I was reading, and I couldn't quite follow this step.  Could someone please help explain this in much greater detail?
If it helps, the author is using integration by parts.
 A: Let
$$
I=-\frac1{2\pi n}\int_0^\infty\frac{d(\cos2\pi nt)}{(x-t)^4}=\int_0^\infty\frac{\sin2\pi nt}{(x-t)^4}dt\,.
$$
Following their cue, let
$$\matrix{
 u=(x-t)^{-4}&\quad
dv=d(\cos2\pi nt)\\\\
du=-4(x-t)^{-5}dt&
 v=\cos2\pi nt
}$$
so that
$$
\eqalign{
-{2\pi n}\,I
&
=\int_{t=0}^{t=\infty}u\,dv
=uv\Bigr|_{t=0}^{t=\infty}
-\int_{t=0}^{t=\infty}v\,du
\\\\
&
=\left.\frac{\cos2\pi nt}{(x-t)^4}\right|_{t=0}^{t=\infty}
+4\int_{t=0}^{t=\infty}\frac{\cos2\pi nt}{(x-t)^5}dt
\\\\
&
=-\frac1{x^4}
+\frac4{2\pi n}\int_{t=0}^{t=\infty}\frac{d(\sin2\pi nt)}{(x-t)^5}
}
$$
which leads to
$$
I = \frac1{2\pi nx^4} - \frac1{\pi^2n^2}
\int_{t=0}^{t=\infty}\frac{d(\sin2\pi nt)}{(x-t)^5}\,.
$$
For the disappearing upper limit, notice that the cosine of anything
is bounded in absolute value by unity, while the denominator blows up,
so that the ratio vanishes.
A: You have
$$ - \frac{1}{{2\pi n}}\int\limits_0^\infty  {\frac{{d\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}}} $$
What you want to do is use integration by parts. This says that
$$\int\limits_a^b {f \cdot g'\left( x \right)dx}  + \int\limits_a^b {f' \cdot g\left( x \right)dx}  = f \cdot g\left( b \right) - f \cdot g\left( a \right)$$
This can be used on your integral, setting
$$\eqalign{
  & \frac{1}{{{{\left( {x - u} \right)}^4}}} = f\left( u \right)  \cr 
  & g'\left( u \right)du = d\cos 2\pi nu \cr} $$
Then we have that
$$\int\limits_0^\infty  {\frac{1}{{{{\left( {x - u} \right)}^4}}} \cdot d\cos 2\pi nu}  - \int\limits_0^\infty  {\frac{4}{{{{\left( {x - u} \right)}^5}}} \cdot \cos 2\pi nudu}  = \mathop {\lim }\limits_{u \to \infty } \frac{{\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}} \cdot  - \frac{{\cos 0}}{{{{\left( {x - 0} \right)}^4}}}$$
It is clear the RHS limit is zero, so let's multiply by $- \frac{1}{{2\pi n}}$ to get
$$ - \frac{1}{{2\pi n}}\int\limits_0^\infty  {\frac{{d\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}}}  = \frac{1}{{2\pi n}}\frac{1}{{{x^4}}} - \frac{4}{{2\pi n}}\int\limits_0^\infty  {\frac{{\cos 2\pi nu}}{{{{\left( {x - u} \right)}^5}}} \cdot du} $$
Now it is just a matter of noticing
$$\cos 2\pi nu\cdot du = \frac{{d\left( {\sin 2\pi nu} \right)}}{{2\pi n}}$$
which gives
$$ - \frac{1}{{2\pi n}}\int\limits_0^\infty  {\frac{{d\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}}}  = \frac{1}{{2\pi n}}\frac{1}{{{x^4}}} - \frac{1}{{{\pi ^2}{n^2}}}\int\limits_0^\infty  {\frac{{d\sin 2\pi nu}}{{{{\left( {x - u} \right)}^5}}}} $$
