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December 19th, 2016, 03:25 PM   #1
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solve this integral?

Hi, I was looking to solve the following integral
\begin{eqnarray}
I=\int dx \frac{1}{x^{2}+a^{2}}\frac{1}{(x+b)^{2}+a^{2}}
\end{eqnarray}
which is the product of two different Lorentzians. Of course, a and b are constants. My initial idea is to spit each Lorentzian into a sum of two terms, like:
\begin{eqnarray}
\frac{1}{x^{2}+a^{2}}=\frac{1}{2ai}\bigg(\frac{1}{ x-ia}-\frac{1}{x+ia}\bigg)
\end{eqnarray}
and the same with the other one, if that might simplify things a bit but I dunno.

Is there any closed formula for this?

Cheers!
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December 19th, 2016, 03:47 PM   #2
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Quote:
Originally Posted by nietzsche View Post
Hi, I was looking to solve the following integral
\begin{eqnarray}
I=\int dx \frac{1}{x^{2}+a^{2}}\frac{1}{(x+b)^{2}+a^{2}}
\end{eqnarray}
which is the product of two different Lorentzians. Of course, a and b are constants. My initial idea is to spit each Lorentzian into a sum of two terms, like:
\begin{eqnarray}
\frac{1}{x^{2}+a^{2}}=\frac{1}{2ai}\bigg(\frac{1}{ x-ia}-\frac{1}{x+ia}\bigg)
\end{eqnarray}
and the same with the other one, if that might simplify things a bit but I dunno.

Is there any closed formula for this?

Cheers!
$\begin{eqnarray}
I=\int dx \frac{1}{x^{2}+a^{2}}\frac{1}{(x+b)^{2}+a^{2}}
\end{eqnarray}= \Large \frac{a \log \left(a^2+b^2+2 b x+x^2\right)-a \log \left(a^2+x^2\right)+b \tan ^{-1}\left(\frac{x}{a}\right)+b \tan ^{-1}\left(\frac{b+x}{a}\right)}{4 a^3 b+a b^3}$
Thanks from nietzsche
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December 19th, 2016, 04:22 PM   #3
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ietzsche, do you now understand how romsek got that? Looks like a candidate for "partial fractions".
Thanks from v8archie

Last edited by Country Boy; December 19th, 2016 at 04:24 PM.
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December 19th, 2016, 04:42 PM   #4
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Quote:
Originally Posted by Country Boy View Post
ietzsche, do you now understand how romsek got that? Looks like a candidate for "partial fractions".
ha, even romsek doesn't understand how romsek got that

I just dumped it into Mathematica and it got spit out.

He just asked for a formula, not how to arrive at it.
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