A Small Taste from My New Book: Episode 5

 

Welcome to Episode 5 of A Small Taste from My New Book. In this episode, we turn our attention to the fascinating world of asymptotic analysis, focusing on Laplace-type integrals and their expansions.

As we explore the integral

for , (Task 1) we’ll see how classical techniques—such as termwise integration and careful control of remainders—lead to precise asymptotic formulas. This approach lays the groundwork for Task 2, where we turn to a more complex contour integral:

Whether you’re interested in the rigorous underpinnings or the practical computations, this episode offers a clear, step-by-step journey through a standard yet elegant result in mathematical analysis.

Task 1

Prove that for

Hint. Make use of the expansion

This is a standard Laplace-type asymptotic expansion. I’ll give a short, fully rigorous derivation using the hint (termwise integration of the finite Taylor polynomial + control of the remainder).

The given expansion is exact for real . It comes from polynomial long division of  by .

Step 1. Split the integrand using given expansion

For every integer  and ,

Multiply by  and integrate on :

No interchange issue here, only finitely many integrals are summed, so linearity immediately justifies it.

Step 2. Evaluate the polynomial integrals

For any integer ,

This is standard use of the Gamma function definition (set ).

Apply this with . The finite sum becomes

So we have the exact identity

where the remainder is

Step 3. Bound the remainder

Since

Thus, for fixed ,

I quote from “The Riemann Hypothesis Revealed”

Asymptotic formulas in mathematics are expressions that approximate the behavior of a function or sequence as the input variable or index approaches a particular limit —often infinity or zero. These formulas help to identify the dominant term in a function’s growth or decay pattern, by revealing how it behaves relative to a simpler, often more intuitive, function in the limit.

For instance, if

we say that  is asymptotic to  as , denoted as . This notation indicates that  and  share the same growth rate as  becomes large, with any difference between them becoming negligible in the limit.

Therefore, if we divide

by

and recall that

it follows that

This shows that  provides an asymptotically accurate approximation for  as  becomes large.

Thus,

This conclusion is also a direct consequence of Watson’s lemma which states that

assuming for   and some ,

(see Example 1 in the paragraph The Gamma Function of The Riemann Hypothesis Revealed).

Task 2

Find the asymptotic expansion of the function

for large .

To tackle this problem, we consider the following

Square root of :

In “The Riemann Hypothesis Revealed” I explain

The square root,  is regarded as the simplest multiple valued function. It has two branches, namely,  and  for  as we can see from the general formula , .

Typically, one uses a keyhole-shaped contour around the negative real axis and deforms it to the upper/lower half-plane. As you traverse from  to  just above the negative real axes,  and , corresponding to the first branch. However, when traversing from  to  just below the real axes, you switch to the second branch which gives to . This careful handling of branches ensures the function remains well-defined along the contour.

I punch the region with two circles  each of fixed radius , around the poles at . This “punching” isolates the singularities, allowing me to apply the residue theorem effectively and account for the contributions from these points.

Let’s start by setting

Cauchy’s theorem ensures

Over

In the limit

I use similar reasoning here as in Example 5 of The Essential Transform Toolkit: by applying the same logic, we see that this integral vanishes as . The same holds for  

Over

Next,

In the limit as

Its modulus satisfies:

Thus,

Contribution from the poles:

We have:

Consequently  

Main contribution:

Finally, on  ,  and on   and  so,

Putting everything together:

I combine the contributions from the residues at the poles and the integrals along the remaining parts of the contour which leads directly to the final identity

Therefore,

Exercise for the Reader

As , the exponential factor  causes the main contribution to the integral to come from values of x near zero.

According to Watson’s lemma, the asymptotic expansion of the integral can be obtained by expanding the algebraic factor

into a power series around  and then integrating each term individually. The rigorous justification for this termwise integration comes from the contour deformation argument developed earlier. Try carrying out this expansion yourself!

Epilogue

The derivation and analysis in this episode showcase the elegance and utility of asymptotic expansions for Laplace-type integrals. By systematically applying polynomial expansions, integrating term by term, and bounding the remainder, we reveal how complex integrals can be approximated with remarkable accuracy for large values of . These results deepen our appreciation for the unity between transform calculus and analytic number theory and underscore the enduring value of classical techniques in modern mathematics.

If you found this exploration insightful, I encourage you to delve further into the detailed discussions in my books, The Riemann Hypothesis Revealed and The Essential Transform Toolkit, available in all Amazon marketplaces. These resources develop the methods presented here with full transparency and step-by-step clarity, making them ideal companions for anyone seeking a deeper understanding of modern mathematical analysis.

And if you read the post and feel lost, that’s the sign to start with my books. They rebuild the foundation clearly and rigorously, without assuming you remember everything from university.

Bonus: analyze the behavior of the integrand along the contour as  and as .