Complement 3: Integral Appendix

Note

This content was crafted by Mr. Ben Madnick.

Note

This integral appendix is modeled after the one in the Blundell and Blundell textbook. An integral appendix is useful in thermodynamics and statistical mechanics because once you are at the thermodynamic limit, distribution functions and probabilities are used to describe various properties of the particles you are studying. These functions are also useful in quantum mechanics, because probabilities and distribution functions become necessary in the regime where classical mechanics fails. One thing worth noting, is that this appendix is here so that, as physicists, we recognize the functions/integrals and use them to simplify otherwise intimidating integrals. For a greater understanding of the derivations of each of the following functions, Introduction to Complex Variables (MATH 4300) is the class to take.

The Gaussian

The Gaussian function ( $\sim e^{-\alpha x^{2}}$ ) turns up in many statistical problems. It is also called the normal distribution or bell curve. The Gaussian is a function of the form $e^{-\alpha x^2}$ where $\alpha$ is a constant. When plotted, as in Figure 1, you can see the bell curve.

Figure 1

From Figure 1, we can see that the function is at a maximum when $x = 0$ . Now since we have described the Gaussian, we can discuss integrating it. We will evaluate using a two-dimensional integral

$\begin{aligned} I^2&=\\ \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy e^{-\alpha(x^2+y^2)} &= \left( \int_{-\infty}^{\infty} dx e^{-\alpha x^2} \right) \left( \int_{-\infty}^{\infty} dy e^{-\alpha y^2} \right) \end{aligned}$

Here $I$ is our desired integral. We can evaluate the left side using polar coordinates:

$\begin{aligned} x &= r\cos{\theta} \\ y &= r\sin{\theta} \\ r^2 &= x^2 + y^2 \\ dxdy &= rdrd\theta \end{aligned}$

Therefore:

$\begin{aligned} I^2 &= \int_{0}^{2\pi} d\theta \int_{0}^{\infty} re^{-\alpha r^2}dr \\ \end{aligned}$

Making the substitution:

$\begin{aligned} z &= \alpha r^2 \\ dz &= 2\alpha rdr \\ \end{aligned}$

This now becomes a simple integral:

$\begin{aligned} I^2 &= 2\pi \left( \frac{1}{2\alpha} \int_{0}^{\infty} e^{-z} dz \right) \\ \end{aligned}$

which can be computed so that:

$I = \sqrt{\frac{\pi}{\alpha}}$

Differentiating both sides with respect to $\alpha$ we find a general formula

$\int_{-\infty}^{\infty} x^{2 n} e^{-\alpha x^{2}} d x=\frac{(2 n) !}{n ! 2^{2 n}} \sqrt{\frac{\pi}{\alpha^{2 n+1}}}, n \geq 0$

Since all of these functions are even, these integrals from $0 \rightarrow \infty$ is just half of this result. This is where they become useful in physics, as this is the velocity distribution for the motion of gas molecules. But what about if $x$ is raised to an odd power? Since the function would be odd, integrating across symmetric bounds means the integral is 0. But that means we still have to consider the integration space from $0 \rightarrow \infty$

$\begin{aligned} I = \int_{0}^{\infty} xe^{-\alpha x^2} dx \\ \end{aligned}$

Making the substitution:

$\begin{aligned} z &= \alpha x^2 \\ dz &= 2x \alpha dx \\ I &= \frac{1}{2\alpha} \int_{0}^{\infty} e^{-z}dz \\ I &= \frac{1}{2\alpha} \end{aligned}$

From here we have a general formula for when $x$ is raised to an odd power. The general formula is

$\int_{0}^{\infty} x^{2 n+1} e^{-\alpha x^{2}} d x=\frac{n !}{2 \alpha^{n+1}}$

The factorial integral and the Gamma Function

Once the class shifts more to the statistical mechanics portion, thermodynamic behavior is studied using mathematical functions and the partition function. The Gamma Function is a useful tool, and an important function to recognize. It’s derivation and use can also be studied in Introduction to Complex Variables. The factorial integral is useful, because once recognized, it simplifies many of the later integrals covered in this appendix and course. The factorial integral is

$n !=\int_{0}^{\infty} x^{n} e^{-x} d x$

It may not be noticed initially, but this function allows one to define the factorial for a non-integer number. Here is where the Gamma Function truly becomes a useful mathematical tool. The traditional definition of the Gamma Function is

$\Gamma(n)=(n-1) !=\int_{0}^{\infty} x^{n-1} e^{-x} d x$

When plotted, the Gamma function looks like Figure 2.

Figure 2

The Gamma Function $\Gamma(n)$ showing the singularities for integer values of $n \leq 0$ . For positive, integer $n$ , $\Gamma(n)=(n-1)!$ 

Riemann Zeta Function

Another function that is involved in many useful integrals is the Riemann zeta function. The Riemann zeta function is defined by

When plotted, the Riemann zeta function appears as

$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^{s}}$

Figure 3

The Riemann zeta function for values $s > 1$ 

Note that the series diverges when $s = 1$ . The analysis of this series can be studied more in Introduction to Complex Variables, but in this class we are interested in its appearance in integrals such as the Bose Integral. The Bose Integral is, as implied from the name, useful in quantum mechanics and statistical mechanics for studying boson gases such as photons. The Bose integral will appear as a consequence of density of states and symmetric states of bosons, but that derivation will be covered in later lectures. The Bose Integral is defined by

$\begin{aligned} I_{B}(n) &= \int_{0}^{\infty} \frac{x^{n}}{e^{x} - 1} dx \end{aligned}$

Evaluating this. We multiply by a factor of 1, in this case, $\frac{e^{-x}}{e^{-x}}$

$\begin{aligned} I_{B}(n) &= \int_{0}^{\infty} \frac{x^{n}e^{-x}}{e^{x}-1} dx \end{aligned}$

The next two steps of this evaluation are explained in greater detail in a complex analysis class. They involve geometric series of complex exponential functions. If you are interested in mathematical functions, it is a good class to take:

$\begin{aligned} I_{B}(n)&= \int_{0}^{\infty} x^{n} dx \sum_{k=0}^{\infty} e^{-(k+1)x} \\ &= \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+1}}\int_{0}^{\infty} y^{n}e^{-y} \\ &= \zeta(n+1)\Gamma(n+1) \end{aligned}$

Thus we have that the Bose integral is

$I_{B}(n)=\int_{0}^{\infty} \frac{x^{n}}{e^{x}-1} d x=\zeta(n+1) \Gamma(n+1)$

Similarly, we also have that

$\int_{0}^{\infty} \frac{x^{n} e^{x}}{\left(e^{x}-1\right)^{2}} d x=n \zeta(n) \Gamma(n)$

The Polylogarithm

The Polylogarithm function, $Li_{n}(z)$ is used in the evaluation of Bose-Einstein and Fermi-Dirac distributions. These distribution functions become important when we begin discussing bosons and fermions. The polylogarithm is defined as

$\begin{aligned} Li_{n}(z) &= \sum_{k=1}^{\infty} \frac{z^k}{k^n} \end{aligned}$

Before we begin showing how this function is useful in solving the distribution functions, we should note that we can write the following geometric progression

$\frac{1}{z^{-1}e^{x}-1} = \frac{ze^{-x}}{1-ze^{-x}} = \sum_{m=0}^{\infty} \left(ze^{-x}\right)^{m+1}$

we then have this integral

$\begin{aligned} \int_{0}^{\infty} \frac{x^{n-1}dx}{z^{-1}e^{x}-1} &= \sum_{m=0}^{\infty} \int_{0}^{\infty} x^{n-1} \left(ze^{-x}\right)^{m+1} \\ &= \sum_{m=0}^{\infty} z^{m+1} \int_{0}^{\infty} x^{n-1}e^{-(m+1)x} dx \end{aligned}$

Making the substitution

$\begin{aligned} u &= (m+1)x \\ du &= (m+1)dx \end{aligned}$

Therefore

$\begin{aligned} x^{n-1} &= \left(\frac{u}{m+1}\right)^{n-1} \end{aligned}$

So the integral becomes

$\begin{aligned} \int_{0}^{\infty} \frac{x^{n-1}dx}{z^{-1}e^{x}-1} &= \sum_{m=0}^{\infty} \frac{z^{m+1}}{(m+1)^{n}} \int_{0}^{\infty} u^{n-1}e^{-u} du \end{aligned}$

This integral should look familiar as it is the Gamma function

$\begin{aligned} \int_{0}^{\infty} \frac{x^{n-1}dx}{z^{-1}e^{x}-1} &= \Gamma(n) \sum_{m=0}^{\infty} \frac{z^{m+1}}{(m+1)^{n}} \end{aligned}$

Making the substitution $k = m+1$ also yields a familiar looking function

$\begin{aligned} \int_{0}^{\infty} \frac{x^{n-1}dx}{z^{-1}e^{x}-1} &= \Gamma(n) \sum_{k=1}^{\infty} \frac{z^k}{k^n} \\ &= \Gamma(n)Li_{n}(z) \end{aligned}$

So we know that this solves the Fermi-Dirac distribution, but what about the Bose-Einstein distribution? The short answer is that it does. If you repeat the preceding process for the BE distribution you will find

$\begin{aligned} \int_{0}^{\infty} \frac{x^{n-1}dx}{z^{-1}e^{x}+1} &= -\Gamma(n)Li_{n}(-z) \end{aligned}$

Combining the two equations we write in general that

$\int_{0}^{\infty} \frac{x^{n-1}}{z^{-1} e^{x} \pm 1} d x=\mp \Gamma(n) L i_{n}(\mp z)$

Performing just a little bit more analysis on the polylogarithm shows that when $\abs{z} \ll 1$ we have that

$Li_{n}(z) \equiv z$

Also we can note that

$Li_{n}(1) = \sum_{k=1}^{\infty} \frac{1}{k^n} = \zeta(n)$

The Dirac Delta function

Here we are going to discuss arguably the simplest function in this appendix. Technically it is a distribution or generalized function. The Dirac delta function is an infinitely high, infinitesimally narrow spike at the origin with an area of 1.

$\delta(x) = \begin{cases} 0, \quad x \neq 0 \\ \infty \quad x = 0 \\ \end{cases}$

and

$\int_{-\infty}^{\infty} \delta(x-a) = 1$

This can be visualized in Figure 4.

Figure 4

This function has some interested properties. What if we want to shift the spike to some arbitrary value $a$ ?. We therefore would have that

$\delta(x-a) = \begin{cases} 0, \quad x \neq a \\ \infty, \quad x = a \\ \end{cases}$

and

$\int_{-\infty}^{\infty} \delta(x-a) dx = 1$

What if the Dirac delta is multiplied by a continuous function $f$ ? We would therefore have

$f(x)\delta(x-a) = f(a)\delta(x-a)$

and

$\int_{-\infty}^{\infty} f(x)\delta(x-a)dx = f(a)$

Imagine that the delta function isolates the value of $f(a)$ and the integral collapses onto that point. The Dirac delta makes many integral simple to evaluate. However, don’t just throw it in to any integral in order to simplify it.

Fourier transforms

While possibly briefly covered in Quantum Physics I and Quantum Physics II, this appendix is here to hopefully give a more thorough explanation of its importance in physics. The Fourier transform allows one to “switch” between time space and frequency space. It is useful for the studying of waves and heat flow problems, which are covered more in a differential equations course. The Fourier transform is defined by

$\begin{aligned} \tilde x(\omega) &= \int_{-\infty}^{\infty} e^{-i\omega t}x(t) \end{aligned}$

This converts the function $x$ from a function of time to a function of frequency. To inverse the transform, we use an inverse transform defined as:

$\begin{aligned} x(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega t} \tilde x(\omega) d\omega \end{aligned}$

One thing to note, be careful about the exponential term. Remember for which transform the exponential term is positive and negative. Both are imaginary but the sign changes between transforms. Also to convert to frequency space the factor of $\frac{1}{2\pi}$ is important and not to be left behind, to ensure proper normalization.

Conclusion

Hopefully this appendix becomes will prove a useful tool not just for Thermodynamics and Statistical Mechanics, but for other math and physics courses. Many of the functions covered in this appendix will also be seen in Astrophysics, Introductory Quantum Mechanics, Complex Variables, and other 4000 level courses at RPI.