Help with simplifying series of hyperbolic integrals

[Strum]

Hello.
I have this function $v(x) = -\sum_{i=1} x^i \sqrt{2}^{i-2} \int_{-\infty}^{\infty} m^{i-1} \cosh(m)^{-4} dm$ which I can not seem to figure out how to simplify.I tried looking at some partial integration but repeated integration of $\cosh$ gives polylogarithms which seemed to only complicate matters. Does anyone have some good tips?
Some observations:
Only uneven $i$ will contribute.
For uneven $i \geq 3$ the integral ($I_i = \int_{-\infty}^{\infty} m^{i-1} \cosh(m)^{-4} dm$) seems to follow the pattern $I_i = a_i\pi^{(i-1)} + b_i\pi^{(i-3)}$, where $a_i$ is going towards $\infty$ with $i$ and the converse is true for $b_i$

 

[fresh_42]

Just an idea. Either you could try to use $\int_{-\infty}^{\infty} \dfrac{dx}{\cosh x} = \pi$ somehow with integration by parts, or what I would do is get rid of the power four, convert the resulting terms into the exponential function and then try to write the integral in terms of the gamma function, and hope to find a nice formula for it.

Another promising ansatz came across my way yesterday: Try to write the equation as a differential equation which might be easier to solve:
https://www.physicsforums.com/threa...-but-not-a-standard-form.932027/#post-5885226
Although this might look differently, in both cases is the gamma function the basic part of the integrand.

 

[Fred Wright]

I have a naive solution to this problem. It may be wrong and I would appreciate comments on errors I may have made. I evaluate the integral for the first term in the series:$$I_1=\int_{-\infty}^{\infty}\frac {dm}{[cosh(m)]^4} = 2^4\int_{-\infty}^{\infty}\frac {e^{4m}dm}{[e^{2m} +1]^4} $$ Let $u=e^m$ and we get$$I_1=2^4\int_0^{\infty}\frac {u^3du}{[u^2 +1]^4}=\int_0^{\infty}K(u)du$$Now make another change of variables: $$v=\frac {1}{[u^2+1]^3}\\dv=-6\frac {u}{[u^2+1]^4}du\\u^2=v^{-1/3} -1\\I_1=\frac{8}{3}\int_0^1u^2dv=\int_0^1(v^{-1/3} -1)dv=\frac{4}{3}$$I now consider the second integral term in the series using the u ($m=log(u)$) change of variables and introduce a parameter, t, for differentiation under the integral sign:$$I_2=\int_0^{\infty}log(tu)K(u)du\\ \frac{dI_2}{dt} =\int_0^{\infty}\frac {\partial log(tu)}{\partial t}K(u)du=\frac{1}{t}\int_0^{\infty}K(u)du=\frac{1}{t}I_1\\I_2=I_1\int \frac {dt}{t}=I_1[log(t)+C]$$Setting $t=1$ we get$$I_2=I_1C$$ and extending this procedure to $I_3$ we find,$$\frac{dI_3}{dt} =\frac{2}{t}I_2\\I_3=2I_1C^2 $$ and by induction,$$I_n=(n-1)!C^{n-1}I_1$$As n goes to infinity the series is clearly divergent and nonphysical unless $C=0$. Therefore the series only has one term with $n=1$.
Salam,
Fred

 

[jasonRF]

Fred Wright,

Using the OPs notation, $I_i = \int_{-\infty}^{\infty} m^{i-1} \cosh(m)^{-4} dm$, for odd $i$ it is clear that the integrand is non-negative for all real $m$, and the only place it is zero is at the origin. So these integrals cannot be zero.

Jason

 

[jasonRF]

Strum,

I believe that contour integration can help you solve this, but the result will be messy. Since your integral is well defined we know
$$ I_n = \int_{-\infty}^{\infty} m^{n-1} \cosh(m)^{-4} dm = \lim_{R\rightarrow \infty}\int_{-R}^{R} m^{n-1} \cosh(m)^{-4} dm$$
I am using $n$ for the index so we don't confuse it with the imaginary number $i$.

Define the function $f(z) = z^{n-1} \cosh^{-4} z$ for compex $z$, and integrate it around a rectangular contour. The lower segment of the contour is along the real axis from $z=-R$ to $z=R$, the right segment of the contour is vertical from $z=R$ to $z=R+i\pi$, the top of the contour is horizontal from $z=R+i\pi$ to $z=-R+i\pi$, and the left side of the contour is from $z=-R+i\pi$ to $z=-R$. I believe the integrals along the right and left sides vanish as $R\rightarrow \infty$, and sum of the top and bottom integrals yields $2 I_n$. Now use the residue theorem, noting that the integrand has a single fourth order pole inside the contour at $z=i\pi/2$. Computing the residue will be messy, but it seems like this approach should work.

EDIT: I was wrong in the above paragraph. The integral along the top of the contour is messy due to the $(z+i\pi)^{n-1}$ term. I think this approach still works, but tue contour integral of $I_5$, for example, will be in terms of $I_3$ and $I_1$.

Good luck.

jason