Standard Integrals

Standard integrals are sets of integrals which we encounter frequently in physics. This is mainly due to the universal nature of the functions involved in such integrals which turn out to be related somehow to the basic equations of the branch of physics under study. For example, while studying atomic and molecular physics we repeatedly encounter exponential functions because in general the atomic wave-functions contain exponentials. Anything things we do with these wave-functions like trying to find out transition probabilities or working out molecular orbitals, etc. we come across integrals of exponentials time and again. It is not convenient to evaluate these integrals every single time. So we try to prepare a set of integrals in terms of which we can usually express more complicated integrals that we encounter while solving problems.

Similarly, in electrostatics we always try to find electric potential by solving Laplace equation for a given system and then using this potential we calculate other physical quantities like field, induced charge, etc. Solutions of Laplace equation can always be expressed in terms of basic functions like Legendre and associated Legendre polynomials (if solved in spherical co-ordinates) and Bessel polynomials (if solved in cylindrical co-ordinates) and potentials for complicated systems are almost alwasys some kind of combination of these functions (polynomials). Hence in electrostatics we always encounter integrals involving these polynomials. So in this case we form a collection of integrals of these polynomials which we can use in solving and simplifying more complicated problems.

In this post I’ll demonstrate a very important and powerful technique called “differentiation under integral sign." This technique comes in handy and many times can easily solve some nasty definite integrals where all other methods would prove hopeless. Yet for some reason I don’t find this addressed in many calculus books. Feynman also mentioned this technique in his “Surely you’re joking Mr. Feynman.” I’ll be using a simplified version of this technique here (interested readers can find a lot of material on this online). Consider the following integral

We can modify this integral a bit by introducing a constant factor in the exponent as shown below and can easily be solved using the substitution method.

 Now the important thing to realize here is that the left-hand side of the above equation is a function of α. Variable ‘x’ acts as a dummy variable here. We could as well have used ‘y’ (or any other letter) in place of ‘x’ and still end up with the same result. However, if we change the value of α, the value of the integral would change, hence, the integral is a function of α. With this in mind, if we differentiate both the sides of the above equation with respect to α, we must have

 Since the integration is over ‘x’ and not α, we can bring the differentiation inside

                                                            

                                                            

Now if we put α = 1, we get

                                                                    

Although we could have easily solved this integral by integration by parts, I chose the example to demonstrate how the technique works. Now if we differentiate the integral second time we get

                                                                    

And again by the same arguments, we end up with
                                                                     

With α = 1, we get

If we wished to solve this by integration by parts, we would have to apply the technique twice and you can imagine how messy the calculations would become if we keep on increasing the power of x. By now you probably would have guessed where I’m going with this. Continuing with the differentiations, it is easy to see that if we differentiate the integral n times, we would get:

                                                       

These are the well known “Gamma Functions.” Now let us consider a more interesting integral

This innocent looking integral defies all of our usual ways of solving integrals. Euler solved it using circular co-ordinates which take about 3-4 lines and there are a couple of other ways of solving it which span over eight or nine pages. The integrand in the above integral is known as the Gaussian and is encountered frequently in statistics and statistical mechanics (when dealing with Normal Distribution, canonical ensemble averages, etc.) and in quantum mechanics (e.g. wave-functions of a harmonic oscillator). Note that a Gaussian is an even function i.e. f(x) = f(-x). Hence we can write the above integral as

Introducing a parameter α just as we did in previous example, we can easily show that

 

The left-hand-side of the above equation is a function of α only. Hence, 

 

 

 

For α = 1, we have

 

Differentiating 2^nd time, we get

 

  

For α = 1, we have

 

Similarly, we can show that by differentiating n times we obtain

 

And for α = 1

 

 

This technique simplifies the calculations significantly while solving definite integrals where other methods would prove too cumbersome. The technique works quite well for a wide range of integrals, especially those involving trigonometric functions, fourier transforms and exponentials. Though like other techniques there are limitations of this technique also. Below is another class of standard integrals frequently encountered in statistical mechanics in the theory of density of states (Debye and Einstein models). These integrals are also related to mathematical objects called Bernoulli Numbers, Riemann-Zeta functions, and Bessel Functions. However, these integrals require different approach than the one that we adopted above. Here try to expand the denominator in Taylor series and then integrate the whole series term by term. This way you will end up with familiar Riemann-Zeta functions which are series of inverse of some power of natural numbers. To see how it works begin with n = 1.