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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.

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I presume $d\cos 2\pi nu$ means $-2\pi n \cdot d\sin 2\pi nu \cdot du$, right? – Pedro Tamaroff Apr 14 '12 at 14:07 You may be right, and that's where I may have made a mistake. Here's why: The step before this one is $\displaystyle\int_0^\infty{\frac{\sin{2 \pi n u} du}{(x-u)^4}}$ – Matt Groff Apr 14 '12 at 14:12
Looks like integratino by parts, but clearly the limits are in $u$. – bgins Apr 14 '12 at 14:31
My main concern is how the term on the left, $\displaystyle\frac{1}{2\pi n x^4}$ shows up. It looks like this is from setting $u=0$. I believe he explains away when $u=\infty$ elsewhere in the paper, and I guess that term is then dropped. – Matt Groff Apr 14 '12 at 14:38
up vote 3 down vote accepted

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.

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I think $du=-4(x-t)^{-5}$ should be $du=-4(x-t)^{-5}dt$ in your integration-by-parts dummy variable array. – Antonio Vargas Apr 14 '12 at 15:03
@AntonioVargas: thanks! – bgins Apr 14 '12 at 15:07

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

$$ - \frac{1}{{2\pi n}}\int\limits_0^\infty {\frac{{d\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}}} = \frac{1}{{2\pi n}}\frac{1}{{{x^4}}} - \frac{1}{{{\pi ^2}{n^2}}}\int\limits_0^\infty {\frac{{d\sin 2\pi nu}}{{{{\left( {x - u} \right)}^5}}}} $$

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