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March 5th, 2018, 03:10 AM   #1
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Solution of this integral?

Let's say we know that $f(x)=f(-x)$, and we want to calculate the Fourier transform of the function $\theta(x)f(x)$, where $\theta(x)$ is the Heaviside step function. Then we define the Fourier transform as:

\begin{eqnarray}

F^{+}(p)=\int_{-\infty}^{+\infty}dxe^{ipx}\theta(x)f(x)=\int_{0}^{ +\infty}dxe^{ipx}f(x)
\end{eqnarray}

and in analogy let's define

\begin{eqnarray}
F^{-}(p)=\int_{-\infty}^{+\infty}dxe^{ipx}\theta(-x)f(x)=\int_{-\infty}^{0}dxe^{ipx}f(x)
\end{eqnarray}

Then it is clear that we have:
\begin{eqnarray}
F^{+}(p)+F^{-}(p)=2\text{Re}[\int_{0}^{+\infty}dxe^{ipx}f(x)] = 2A(p)\\
F^{+}(p)-F^{-}(p)=2\text{Im}[\int_{0}^{+\infty}dxe^{ipx}f(x)]= 2B(p)
\end{eqnarray}
We are interested in the calculation of $F^{+}(p)$, BUT WE ONLY KNOW A, NOT B! This means another equation is necessary in order to compute $F^{+}(p)$.

In the concrete example I am working on, I have found the following:
\begin{eqnarray}

\int_{0}^{+\infty}dx\cos(px)\frac{1}{\cos(a)-\cosh(x)} = A(p)
\end{eqnarray}
where $a>0$ is a constant, and A(p) is well known in this form, so that we have defined
\begin{eqnarray}

f(x)=\frac{1}{\cos(a)-\cosh(x)}
\end{eqnarray}


. Thing is, I am having difficulties to compute the imaginary part of the FT in the half line, that is, the term
\begin{eqnarray}

\int_{0}^{+\infty}dx\sin(px)\frac{1}{\cos(a)-\cosh(x)} = ?
\end{eqnarray}

Is there a way to compute this integral?? Any other use of the symmetry of the function f(x)?

Last edited by skipjack; March 7th, 2018 at 07:42 AM. Reason: to change cosh(a) to cos(a)
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March 6th, 2018, 10:30 AM   #2
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Ok I have an update, where I have performed the following calculation by using Wolfram Mathematica:

\begin{align*}

F^{+}(p)&=-\frac{1}{2\pi}\int_{0}^{+\infty}dx e^{ipx}\frac{\sin(2a)}{\cos(2a)-\cosh(x)} \\
&=\frac{1}{4\pi}\int_{0}^{+\infty}dx e^{ipx}\frac{\sin(2a)}{\sin(a + i(x/2))\sin(a-i(x/2))}
\end{align*}

and by use of further simplifications from the WA solution, I get

\begin{eqnarray}

F^{+}(p)=-\frac{1}{\pi}\sum_{k=0}^{+\infty}\frac{\sin(2a(k+1 ))}{1-ip + k}
\end{eqnarray}

Is there a way to write this series in a closed form? Also, given the fact that $F^{+}(p)$ must be real valued, can we just take the summing part of the purely imaginary side equal to 0? By this, I mean dividing the series above into real and imaginary parts, but knowing that the series of the imaginary part must be identically zero, which I don't think is something immediate to prove, any ideas welcome please!!

Last edited by skipjack; March 8th, 2018 at 01:30 AM.
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March 6th, 2018, 11:10 AM   #3
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Hi nietzsche,

Since $\theta(x)$ is the Heaviside step function, I think what you are calling $F^-(p)$ is 0 since $\theta(x)=0, \text{ for } x <0$. In your definition of $F^-(p)$ you take $\theta(-x)f(x)$ instead of $\theta(x)f(-x)=\theta(x)f(x) = 0, \ x<0$ .

I do not know if this was intentional or not but as a result I think that you have multiplied $f(x)$ by 1 for all $x$ instead of multiplying by $\theta(x)$.

So it appears that you cannot use symmetry since the result of the multiplication by $\theta(x)$ will give neither an even nor odd function. So there will be the pesky sin integral.

However, for the concrete example that you gave, I think that it is worse. I don't think that this function even has a Fourier Transform because my textbook says that $f(x)$ needs to be absolutely integrable:

$\displaystyle \large \int_{-\infty}^{\infty} |f(x)| \ dx \ < \ \infty$

Since $a>0$ is given, let's pick $a=2$. W|A says that $f(x)$ does not converge but it says it ran out of computing time, so maybe it does converge.

Hmm, now I see your update but it seems that $f(x)$ has changed to a constant, depending upon the constant $a$ . Is this correct? I may have missed something or misinterpreted your result.
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March 6th, 2018, 01:57 PM   #4
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Hi jks,

Thanks for your reply, the definitions of $F^{+}(p)$ and $F^{-}(p)$ are correct as far as I know, they just represent the Fourier transforms of the function f(x) ONLY IN THE HALF LINES $(-\infty,0)$ and $(0,+\infty)$. That is why $F^{+} + F^{-}$ gives the real part of the Fourier transform of the function f(x), which in my case is well known.

Respect to the specific function I used, all Fourier transforms are well defined, whether or not that results in a closed analytical expression is a different question. According to WA, the integral of that function with the complex exponential gives a combination of two hipergeometric functions, that can be manipulated to give the series I wrote above; how to proceed from there is where I am really stucked, any suggestions are welcome!!
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March 6th, 2018, 02:30 PM   #5
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Quote:
Originally Posted by jks View Post
Hi nietzsche,


$\displaystyle \large \int_{-\infty}^{\infty} |f(x)| \ dx \ < \ \infty$
There is an error in my initial integral form, the constant in the denominator should be $\cos(a)$, not $\cosh(a)$ which is unbounded. One can pass to the hiperbolic cosine just by considering that $\cos(a)=\cosh(ia)$, sorry about this!

Last edited by nietzsche; March 6th, 2018 at 02:32 PM.
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March 6th, 2018, 02:35 PM   #6
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Quote:
Originally Posted by jks View Post
Hmm, now I see your update but it seems that $f(x)$ has changed to a constant, depending upon the constant $a$ . Is this correct? I may have missed something or misinterpreted your result.
No, $f(x)$ is not a constant, it is simply reestructured in terms of csc( aix + b)csc(b-aix), there is still an $x$ dependence! otherwise the integral would be trivial
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March 6th, 2018, 10:37 PM   #7
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Ok,I think I see what you are doing. I am not familiar with your specific problem but I question why you think that $F^+(p)$ is purely real.

If you look at the way it is defined, $F^{+}(p)=\int_{0}^{ +\infty}dxe^{ipx}f(x)$ then it is evident that from 0 to $+\infty, f(x)$ has neither even nor odd symmetry. So $F^+(p)$ will have both real and imaginary parts.

Also, if you add your equations reproduced below:

$\displaystyle \begin{eqnarray}
F^{+}(p)+F^{-}(p)=2\text{Re}[\int_{0}^{+\infty}dxe^{ipx}f(x)] = 2A(p)\\
F^{+}(p)-F^{-}(p)=2\text{Im}[\int_{0}^{+\infty}dxe^{ipx}f(x)]= 2B(p)
\end{eqnarray}$

You get:

$\displaystyle \begin{align*}
2F^+(p)&=2A(p)+2iB(p) \\
F^+(p)&=A(p)+iB(p)
\end{align*}$

Does this seem correct?
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March 7th, 2018, 03:58 AM   #8
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Quote:
Originally Posted by jks View Post
Ok,I think I see what you are doing. I am not familiar with your specific problem but I question why you think that $F^+(p)$ is purely real.
So, the function $f(x)=f(-x)$ by definition, like
\begin{eqnarray}
f(x)=\frac{\sin(2a)}{\cos(2a)-\cosh(x)}=f(-x)
\end{eqnarray}

We just want the integral of this function times $e^{ipx}$ for $x\in [0,+\infty)$
Quote:
Originally Posted by jks View Post
If you look at the way it is defined, $F^{+}(p)=\int_{0}^{ +\infty}dxe^{ipx}f(x)$ then it is evident that from 0 to $+\infty, f(x)$ has neither even nor odd symmetry. So $F^+(p)$ will have both real and imaginary parts.
That is absolutely correct, I just realised now I might have messed up with that property of Fourier transforms, but $F^{+}(p)$ is not a proper FT.That is why when I get the series

\begin{eqnarray}
\sum_{n=0}\frac{\sin(2a(n+1))}{1-ip+n}
\end{eqnarray}
there is an imaginary part contribution. Thing is from equation

\begin{eqnarray}

F^{+}(p)=A(p)+iB(p)
\end{eqnarray}
we still dont know B(p) in closed form, although I have managed to find the solution of the integral with the cosine part (A(p)) when instead of $\cos(a>0)$ we have $\cosh(a>0)$ in the denominator, which seems to be a Cauchy Principal Value integral...
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March 7th, 2018, 08:33 AM   #9
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UPDATE:

I just came with an expression in terms of Lerch zeta functions for the series above. According to my calculation, the total function F^{+}(p) now reduces to:
\begin{eqnarray}
F^{+}(p)=\frac{1}{2\pi i}\bigg(e^{2i\sigma}L(\sigma/\pi,z,1) - e^{-i2\sigma}L(-\sigma/\pi,z,1)\bigg)
\end{eqnarray}
where we have defined
\begin{eqnarray}
z &=& 1-ip\nonumber\\
L(\lambda,z,s) &=&\sum_{n=0}^{+\infty}\frac{e^{i2\pi\lambda n}}{(n+z)^{s}}
\end{eqnarray}

Is there a way to reduce this even further? (It is the first time I encounter these functions, but I am quite happy to see this beautiful result )
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