Integrating a Function with Poles

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In this example we want to use Cauchy’s Integral Theorem to integrate a function that has actual singularities (poles) in it. We will actually encounter such a function if we seek to calculate the pressure exerted by a photon gas, the pressure that blackbody radiation exerts upon the walls of a cavity in which it is trapped.

In a cavity of volume V a photon of energy Ei contributes to the pressure of the photon gas an amount p=- Ei/ V, so the mean pressure from all of the photons conforms to the description

(Eq’n 1)

in which <ni> represents the mean number of photons in the i-th energy state. Let the cavity have the form of a cube whose sides have length L. The allowed energy states in that cavity conform to the description

(Eq’n 2)

in which Ci represents a constant. Thus we have

(Eq’n 3)

We can now rewrite Equation 1 as

(Eq'n 4)

in which <E> represents the average of the total energy in the photon gas and <u> represents the average energy density in the radiation field. Planck’s blackbody radiation theorem describes that energy density as

(Eq’n 5)

in which η= ħω/kT and T represents the absolute temperature of the cavity walls. Now we want to evaluate the integral in that equation.

To that end we rewrite the integrand as

(Eq’n 6)

We then get our integral as

(Eq’n 7)

That equation gives us I0=π4/15. To prove and verify that statement we need only evaluate the infinite sum.

That sum represents the Riemann zeta function of index four,

(Eq’n 8)

We can’t calculate that sum directly, because we can’t hope to add up an infinite set of terms directly, so we need to be sneaky. We want to convert that sum into an integral that corresponds to the sum; more specifically, we want the integral to extend over the domain 1≤ x<∞ and yet contribute nonzero increments to its Riemannian sum only at the integer values of x. Then we can extend the domain of integration over the entire complex plane and exploit Cauchy’s Integral Formula. To achieve that end we apply a periodic Dirac Delta to our function and rewrite our sum as

(Eq’n 9)

Thus we convert the Riemann zeta function from a sum of discrete elements into an integral. The integrand of that integral only has nonzero values at integer values of x, values for which the sine in the argument of the Dirac Delta goes to zero. The cosine gives comes into play as part of the differential associated with the functional-argument Dirac Delta. Next we extend the limits of the integration from 1≤ x<∞ to -∞ <x<∞ and get

(Eq’n 10)

Because we have used an even power of x, we have the integral evaluated from minus one to minus infinity equal to the integral in Equation 9, so the sum of those integrals equals 2ζ(4). But by extending the limits on the integration to cover the entire number line we have also given ourselves a singularity at x=0.

To make that integral susceptible to Cauchy’s Integral Theorem we must extend its domain out onto the complex plane. So we replace x with z=x+iy to extend our function over the entire complex plane and then use the trigonometric and hyperbolic identities to write the argument of the Dirac Delta as

(Eq’n 11)

Because cosh(πy) never takes a value less than one and sinh(πy) can only go to zero when y=0, that function only goes to zero when y=0. So now we know that extending our integral over the complex plane doesn’t give us any more points where the Dirac Delta takes nonzero values and, thus, doesn’t add any more terms to the sum in our zeta function. Cauchy’s Integral Theorem tells us that if we integrate that function on a contour that encloses the entire complex plane, the result must equal zero. Of course, the complex plane has no proper boundary: with infinite radius the putative boundary cannot represent an actual specific place where the complex plane ends. For the purpose of using Cauchy’s Integral Theorem we assume a circle of arbitrarily large radius, centered on the origin of the plane, and then take the limit of the integral as we let the radius of that circle increase endlessly.

But in order to invoke Cauchy’s Integral Theorem, we must extend the domain of the integration over the entire complex plane. That fact necessitates that we extend the domain of the integration from between one and infinity to between minus infinity and plus infinity. That extension will add extra terms to our sum, those that come from the negative integers plus a singularity at x=0. Because we have used an even function, the sum of the negative-integer terms equals the sum of the positive-integer terms, so if we add up all of the terms in the new domain, we need only divide that sum by two to get the correct value of the zeta function. The singularity at the origin of the number line will give us the means to achieve that calculation.

Cauchy’s Integral Theorem (also known as the Cauchy-Goursat theorem) tells us that if we have a function that remains analytic on a closed contour and on all points enclosed within that contour, then the integral of that function on the contour equals zero. In this case we want the contour to coincide with the boundary enclosing the entire complex plane. Of course, the complex plane has no proper boundary: with infinite radius, the putative boundary cannot represent an actual, specific place where the complex plane ends. For the purpose of using Cauchy’s Integral Theorem we assume a circle of arbitrarily large radius and take the limit of the integral as the radius of that circle increases endlessly.

Cauchy’s Integral Theorem comes from applying Green’s theorem,

(Eq’n 12)

to analytic functions on the complex plane. Cauchy equates the line integral of an analytic complex-valued function on the boundary of a region to the area integral of that function’s "curl" on the region itself, uses the Cauchy-Riemann conditions to zero out the curl, and thereby zeroes out the line integral. In essence Cauchy made a feint at creating the complex analysis analogue of Ampere’s law, but produced instead a simple version in which we preserve the zeroness of the line integral by cutting out any singularities in the region within the line C bounding the region R. But if the function that we want to integrate satisfies a simple criterion, we can apply Jordan’s lemma and develop a true pure-mathematics version of Ampere’s law.

If we have a function, f(z), whose value diminishes faster than 1/|z| as the magnitude of z increases and if we have a semicircle CR of radius R whose every point conforms to z=Rexp[iθ], then we have as true to mathematics the fact that

Eq’n 13)

That equation expresses Jordan’s lemma and it looks very much like Cauchy’s Integral Theorem. It differs only in specifying that CR represent a semicircle centered on the origin of the complex plane. We can close that contour by adding to it the real-number axis, thereby creating a contour that completely encloses the upper half of the complex plane or the lower half of the complex plane. If we have an integral that we need to evaluate over the entire real-number axis, we can exploit the residue theorem, using Jordan’s lemma to eliminate the part of the contour off the real axis; that is, for some f(z) we have

(Eq’n 14)

The example that we are pursuing has a pole at the point z=0. All of the other singularities on the real axis, also based on the Dirac Delta, when integrated add up to twice the value of the zeta function. So we have, by Equation 14,

(Eq’n 15)

Thus we need to calculate

(Eq’n 16)

In that equation the minus sign comes in because the rest of the integration on the real-number axis yields a positive number, which this integral, in its raw form, must compensate in order to yield the necessary zero. And in jumping from the first to the second line of that equation I have played a number of mathematical tricks. First I extended the domain of the integration over the complex plane (something we had already done implicitly by invoking Jordan’s lemma) and replaced the variable x with z=x+iy (treating y as an implicit, hidden variable that had been present all along). Thus, in the first integral we integrate across the singularity and in the second integral we integrate around it, as Cauchy taught us to do. Next I replaced the Cartesian representation of the complex plane with the corresponding polar representation (z=rexp(iθ) and dz=izdθ), which necessitated dividing the Dirac Delta by two pi eye times its argument (noting that the square root of minus one comes from using the complex plane rather than the standard Cartesian plane of geometry). The differential element of area, (πdx)(πdy)=π2rdθdr, reflects my choice to apply a Cauchy integration (assuming that r is held constant during the integration around the contour) and then an integration over radial distance along the real-number axis.

To convert the integral of Equation 16 into Cauchy integrable form we exploit the fact that

(Eq’n 17)

Multiplying that equation by the reciprocal fourth power of z and by the Dirac Delta gives us our integrand as an infinite series, of which only the 1/z term survives the Cauchy integration. Thus we rewrite Equation 16 as

(Eq’n 18)

Carrying out the longitudinal integration around a minuscule circle gives us

(Eq’n 19)

Although the argument of the Dirac Delta is a function of θ, the Dirac Delta has a constant value (zero) everywhere except at z=0, where θ becomes an ignorable coordinate, so the Dirac Delta comes through the integration with respect to θ unchanged. We can transform the Dirac Delta of sin(πz) into the Dirac Delta of πr under the condition that θ=0. That move leads us to

(Eq’n 20)

because at r=0, cos(πr)=1.

At last we can substitute that result into Equation 7 and thence into Equation 5 to get

(Eq’n 21)

And, of course, by Equation 4, the pressure equals one third of that.

For temperatures less than the melting point of carbon, that pressure will come out very small. We would expect the container to melt or vaporize before the radiation it contains exerts a pressure capable of measurably deforming the material. But, according to Sir Arthur Eddington and other astrophysicists, the pressure exerted by the radiation trapped within them prevents the stars from collapsing into planet-sized dwarfs. Long before it reaches us, the light coming from the sun has done the tremendous work of holding up the sun, keeping it large enough to make it suitable for warming a life-bearing planet. In that way astrophysicists deduce the rules that govern the operation of the Universe.

gabh

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