My Math Forum Solution of this integral?

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 March 5th, 2018, 02:10 AM #1 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 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 06:42 AM. Reason: to change cosh(a) to cos(a)
 March 6th, 2018, 09:30 AM #2 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 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 12:30 AM.
 March 6th, 2018, 10:10 AM #3 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 626 Thanks: 90 Math Focus: Electrical Engineering Applications 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.
 March 6th, 2018, 12:57 PM #4 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 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!!
March 6th, 2018, 01:30 PM   #5
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Quote:
 Originally Posted by jks 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 01:32 PM.

March 6th, 2018, 01:35 PM   #6
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Quote:
 Originally Posted by jks 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

 March 6th, 2018, 09:37 PM #7 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 626 Thanks: 90 Math Focus: Electrical Engineering Applications 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?
March 7th, 2018, 02:58 AM   #8
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Quote:
 Originally Posted by jks 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 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...

 March 7th, 2018, 07:33 AM #9 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 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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