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I would like to solve the following integral:

$$ I_0 (a,b)= \int_0^1 dx\int_0^{1-x} dz \frac{1}{a z (z-1)+a x z + x(1-b)}$$ in the limit where $b$ is small ($a$ and $b$ are positive constants).

What is the best way to study this asymptotic limit? I have tried to use Mathematica, but it gives some very weird results.

Thanks for any help.

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  • $\begingroup$ Hi JSchwinger, first, is this integral finite at all? I didn't check it, but one doesn't see it at the first glance. Second, you want to study its behavior for a>0 fixed and b-> 0 with b>0, right? Asymptotic expansions are often a little bit tricky, so I doubt that Mathematica can do it. $\endgroup$ – Karl Oct 29 '14 at 6:36
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For now we ignore the justification for switching the order of integration and solve an easier problem with the help of Mathematica. The region of integration is the triangle bounded by the $x$ and $z$ axes and the line $x + z = 1$. When we switch the order of integration we have the integral

$$J(a,b) = \int_0^1 \int_0^{1-z} \frac{dx dz}{az(z-1)+axz+x(1-b)}$$

The denominator is $0$ whenever

$$x_0 = \frac{az(1-z)}{az+1-b}$$

so we need to calculate an improper integral first. We have

$$\int_0^{x_0-\epsilon} \frac{dx}{az(z-1)+axz+x(1-b)} = \frac{\log((-1+b-az)\epsilon)-\log(az(z-1))}{az+1-b}$$

and since this is not finite as $\epsilon$ goes to $0$, we have to concede that the improper integral does not converge. Similarly

$$\int_{x_0+\epsilon}^{1-z} \frac{dx}{az(z-1)+axz+x(1-b)} = \frac{\log((1-b)(1-z))-\log((1-b+az)\epsilon)}{az+1-b}$$

is not finite as $\epsilon$ goes to $0$. However, we can calculate the principal value by putting the two integrals together and let the limit of integration approach $x_0$ at the same rate on both sides

$$P.V.\int_0^{1-z} = \frac{\log((1-b)(1-z))-\log(az(1-z))}{1-b+az}$$

Integrating again with respect to z we have

$$P.V.J(a,b) = \int_0^1\frac{\log((1-b)(1-z))-\log(az(1-z))}{1-b+az} dz = \frac{1}{a}\left(\log\left(\frac{a}{1-b}\right)\log\left(\frac{1-b}{1+a-b}\right)-Li_2\left(-\frac{a}{1-b}\right)\right)$$

We can expand this as

$$-\frac{1}{a}(\log(a)\log(1+a)+Li_2(-a))-\frac{\log(a)}{1+a}b-\frac{1+a+(2+a)\log(a)}{2(1+a)^2}b^2+O(b^3)$$

When $b$ is small we are left with

$$-\frac{1}{a}(\log(a)\log(1+a)+Li_2(-a))$$

The question is - were we justified in exchanging the order of integration in the first place? I'm not sure if this helps.

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  • $\begingroup$ Than you, @Aleks. That helps a lot! Do you think that the exchanging of the order of integration could raise any problems? $\endgroup$ – JSchwinger Nov 2 '14 at 13:03
  • $\begingroup$ Unfortunately I don't know the answer to that in this case. All the theorems I know about exchanging the order of integration have a continuous integrand which is not the case here. You'd need a theorem discussing changing the order of integration for principal values. $\endgroup$ – Alexander Vlasev Nov 2 '14 at 20:12
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EDIT added later

The integrand's denominator is zero along a certain hyperbola that passes through the overall region of integration. So as written, the integral is improper. I wonder if you mean for $a$ to be a small negative constant instead. Then the function is defined inside the region, and as $a\to0^-$, this hyperbola straightens out to the $z$-axis, which is one of the region's boundary curves. Proceeding assuming that $a<0$ and $b>0$:

End EDIT


The integrand's denominator is linear in $x$, so let's try integrating with respect to $x$ first. Note that the region of integration is he triangle with vertices at $(x,z)=(0,0),(1,0),(0,1)$, so the limits on the integrals can be swapped.

$$\begin{align} &\int_{z=0}^{z=1}dz\int_{x=0}^{x=1-z}dx\frac{1}{a z (z-1)+a x z + x(1-b)}\\ &=\int_{z=0}^{z=1}dz\int_{x=0}^{x=1-z}dx\frac{1}{a z (z-1)+ x(1-b+az)}\\ &=\int_{z=0}^{z=1}dz\frac{1}{1-b+az}\left[\ln\left|a z (z-1)+ x(1-b+az)\right|\right]_{x=0}^{x=1-z}\\ &=\int_{z=0}^{z=1}dz\frac{1}{1-b+az}\left[\ln\left|a z (z-1)+ (1-z)(1-b+az)\right|-\ln\left|a z (z-1)\right|\right]\\ &=\int_{z=0}^{z=1}dz\frac{1}{1-b+az}\left[\ln\left| \left(1-b\right)\left(1 - z\right)\right|-\ln\left|a z (z-1)\right|\right]\\ &=\int_{z=0}^{z=1}dz\frac{1}{1-b+az}\left[\ln\left|\frac{b-1}{az}\right|\right]\\ &=\int_{z=0}^{z=1}dz\frac{\ln\left|b-1\right|}{1-b+az}-\int_{z=0}^{z=1}dz\frac{\ln|az|}{1-b+az}\\ &=\left[\ln\left|b-1\right|\frac1a\ln\left|1-b+az\right|\right]_{z=0}^{z=1}-\frac{1}{a}\int_{u=0}^{u=a}du\frac{\ln|u|}{1-b+u}\\ &=\frac{\ln\left|b-1\right|}{a}\ln\left|1+\frac{a}{1-b}\right|-\frac{1}{a}\int_{u=0}^{u=a}du\frac{\ln|u|}{1-b+u}\\ \end{align}$$

This last integral is not one that has an elementary antiderivative. Its antiderivative involves a function called the dilogarithm, $\operatorname{Li}_2$, which could be thought of as a function that was invented just to give a name to a certain antiderivative that was out of reach within the world of elementary functions.

If you want to go further, $$\int\frac{\ln u}{A+u}\,du=\operatorname{Li}_2\left(-{\frac{u}{A}}\right)+\ln(x)\ln\left(1+\frac{x}{A}\right)+C$$ and also because one limit of integration is $0$, this integral is improper, and should be computed using a limit as $t\to0^+$ of $\int_t^{a}$.

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