Calculus Calculus Math Forum

 March 5th, 2018, 03:10 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 07:42 AM. Reason: to change cosh(a) to cos(a) March 6th, 2018, 10:30 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 01:30 AM. March 6th, 2018, 11:10 AM #3 Senior Member   Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 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, 01:57 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, 02:30 PM   #5
Member

Joined: Dec 2016
From: -

Posts: 62
Thanks: 10

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 02:32 PM. March 6th, 2018, 02:35 PM   #6
Member

Joined: Dec 2016
From: -

Posts: 62
Thanks: 10

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, 10:37 PM #7 Senior Member   Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 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, 03:58 AM   #8
Member

Joined: Dec 2016
From: -

Posts: 62
Thanks: 10

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, 08:33 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 ) Tags integral, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shy Calculus 2 October 31st, 2014 12:39 AM cuasimodo Calculus 0 April 21st, 2014 12:35 AM mathLover Calculus 5 May 20th, 2013 10:08 PM mathLover Calculus 10 December 3rd, 2012 11:22 AM debad Calculus 9 July 1st, 2012 03:18 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      