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Many recent questions have been asked here similar to this integral

$$\int_{-\infty}^\infty \frac {dx}{\cos x + \cosh x} = 2.39587\dots$$

whose "closed form" I cannot seem to figure out. I have tried contour integration, but the sum of the residues got rather nasty.

Can someone perhaps evaluate this integral?

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    $\begingroup$ Is there reason to expect that it has a closed form? $\endgroup$ Commented Feb 6, 2013 at 4:18
  • $\begingroup$ In terms of having $\cos x+\cosh x$ in denominator here are somewhat similar and another problem $\endgroup$
    – Maesumi
    Commented Feb 6, 2013 at 5:25
  • $\begingroup$ @Joseph Yes, because it has been posted, albeit in more complicated forms, in two other questions I have read, both of which seem to imply a closed form exist. $\endgroup$
    – Argon
    Commented Feb 6, 2013 at 22:23

3 Answers 3

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Simple computation yields $$ \cos(z)+\cosh(z)=2\cos(\alpha z)\cosh(\alpha z)\tag{1} $$ where $\alpha=\frac{1+i}{2}$. The zeros of $\cos(z)+\cosh(z)$ are at $$ z_k^\pm=\left(k\pi+\frac\pi2\right)(1\pm i)\tag{2} $$ and $$ \operatorname*{Res}_{\ \ z=z_k^\pm}\frac1{\cos(z)+\cosh(z)} =\frac{1\pm i}{2}(-1)^{k+1}\mathrm{sech}\left(k\pi+\frac\pi2\right)\tag{3} $$ The residues in the upper half-plane are $z_k^+$ where $k\ge0$ and $z_k^-$ where $k\lt0$. Pairing the residues at $z_k^+$ and $z_{-k-1}^-$ yields the sum of the residues in the upper half-plane to be $$ -i\sum_{k=0}^\infty(-1)^k\mathrm{sech}\left(k\pi+\frac\pi2\right)\tag{4} $$ Therefore, we can use contour integration to get $$ \begin{align} \int_{-\infty}^\infty\frac{\mathrm{d}x}{\cos(x)+\cosh(x)} &=2\pi\sum_{k=0}^\infty(-1)^k\mathrm{sech}\left(k\pi+\frac\pi2\right)\\[6pt] &\doteq2.3958786339145620925\tag{5} \end{align} $$

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  • $\begingroup$ Funny, I got something like this, concluded at the time that it wasn't good enough, and have yet to see anything better, including another angle I am trying. (+1) $\endgroup$
    – Ron Gordon
    Commented Feb 6, 2013 at 11:24
  • $\begingroup$ @rlgordonma: It isn't as pretty as a closed form, but $\mathrm{sech}(x)$ decays exponentially, so the sum converges pretty quickly. $\endgroup$
    – robjohn
    Commented Feb 6, 2013 at 11:30
  • $\begingroup$ I agree. Compare this with what I was trying - I did away with exponentials, which in hindsight is looking for trouble unless you strike gold. Still, I posted because maybe someone could tell me why trouble occurs when switching the order of summation and integration - or if I made a major error somewhere. $\endgroup$
    – Ron Gordon
    Commented Feb 6, 2013 at 11:32
  • $\begingroup$ What contour are you integrating over when using the residue theorem? You need to be very careful here, to avoid the poles and choose the contour so that the integral over the ''extra part'' tends to $0$. $\endgroup$
    – mrf
    Commented Feb 6, 2013 at 16:18
  • $\begingroup$ Beat me hands down! :) $\endgroup$
    – user17762
    Commented Feb 6, 2013 at 16:41
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We have \begin{align} I & = \int_0^{\infty} \dfrac{dx}{\cos(x) + \cosh(x)}\\ & = \int_0^{\infty} \dfrac{\text{sech}(x)dx}{1 + \text{sech}(x)\cos(x)}\\ & = \sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx \end{align} where \begin{align} \int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx & = 2^{k+1}\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx \end{align} Now we have $$\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx = \int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx$$ Further, $$\dfrac1{(1+e^{-2x})^{k+1}} = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l e^{-2lx}$$ Hence, $$\int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx$$ Hence, \begin{align} \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx & = \int_0^{\infty} e^{-(2l+k+1)x}\left(\dfrac{e^{ix}+e^{-ix}}2 \right)^k dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{imx}e^{-i(k-m)x}dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{2imx-ikx}dx\\ & = \dfrac1{2^k} \sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \end{align} Hence, $$I = 2 \sum_{k=0}^{\infty}(-1)^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \right)$$ The integral you are after is twice $I$ and hence $$\color{blue}{4 \sum_{k=0}^{\infty}(-1)^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \right)}$$ You can truncate the two infinite sums (since both are alternating sums) to get arbitrary accuracy.

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  • $\begingroup$ I think that you must warn the reader (mainly the OP) about the validity of the formal manipulations you perform along the answer. To me it seems more difficult to justify them that the result itself. $\endgroup$ Commented Feb 6, 2013 at 4:38
  • $\begingroup$ That looks as nice as a whole, sunny and beautiful friday morning wasted talking to a life-insurance agent...wow! I'm not sure whether that awful-looking triple sum can even be decently approximated, let alone calculated accurately...can it? $\endgroup$
    – DonAntonio
    Commented Feb 6, 2013 at 5:01
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    $\begingroup$ @Marvis: how on earth does the integral have a dependence on $x$ when $x$ is the integration variable? I expect a number. $\endgroup$
    – Ron Gordon
    Commented Feb 6, 2013 at 9:18
  • $\begingroup$ @rlgordonma Typo corrected. $\endgroup$
    – user17762
    Commented Feb 6, 2013 at 16:42
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OK, this is an especially tough integral. Mathematica gives nothing analytical, nor could I find a form in Gradshteyn & Rhyzik that covered this integral. That said, it is still worth looking into what can be done, and I think I have some sort of an advance over what I've seen so far.

Write the integral as

$$I = 4 \int_0^{\infty} dx \: \frac{e^{-x}}{1+2 e^{-x} \cos{x} + e^{-2 x}} $$

Substitute $y=e^{-x}$, $dy=-y dx$, $x=-\log{y}$ and get

$$I = 4 \int_0^1 \frac{dy}{1+2 y \cos{(\log{y})} + y^2}$$

This is still a tough integral, but maybe some of you recognize that the integrand is a generating function for the Chebyshev polynomial of the second kind $U_n(x)$. That is, we may write

$$\frac{1}{1+2 y \cos{(\log{y})} + y^2} = \sum_{n=0}^{\infty} U_n(\cos{(\log{y})}) y^n$$

Now we have

$$I = 4 \int_0^1 dy \: \sum_{n=0}^{\infty} y^n U_n(\cos{(\log{y})}) $$

I am not yet convinced that the order of summation and integration may be reversed. (Upon doing so, the resulting series diverges as far as I can see, which is too bad because the results were coming out nice.)

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  • $\begingroup$ Ron, why do you ask whether sum and integral can be switched? Of course they can, summation is associative. $\endgroup$ Commented Oct 9, 2014 at 19:20
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    $\begingroup$ @PeterHalburt finite sums and integrals can be switched. Since infinite sums are limits, as are integrals, care must be taken. Sometimes switching the order of limits affects the outcome. $\endgroup$
    – robjohn
    Commented May 30, 2015 at 1:55

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