How to integrate convergent integrals using divergent integrals

Methods abound in evaluating convergent integrals, for example, by substitution or by differential equations. In this post I show how divergent integrals may be used in evaluating convergent integrals by applying our recent results in the complex contour integral representations of the finite part of divergent integrals found here and here. To be concrete, let us consider the integral
\begin{equation}
\int_0^{\infty} \frac{e^{-a x} – e^{-b x}}{x^{3/2}} \mathrm{d}x = 2 (\sqrt{b}-\sqrt{a}) \sqrt{\pi}.
\end{equation}

Let us attempt to evaluate the integral by distributing the integral over the two terms. We obtain the difference of two integrals,
\begin{equation}
\int_0^{\infty} \frac{e^{-a x}}{x^{3/2}} \mathrm{d}x – \int_0^{\infty} \frac{e^{-b x}}{x^{3/2}} \mathrm{d}x
\end{equation} which are both divergent.

Can we make sense of the divergent integrals to obtain the value of the given integral? Dealing with this question requires us to properly interpret the divergent integrals involved. The complication in dealing with divergent integrals is that there is no single rule of assigning values to them. A divergent integral can be assigned meaningful value by means of, say, analytic continuation, Caesaro limits, regularization methods, distributional methods, or finite part integration. These generally assign different values to the divergent integral. Here we chose to assign a given divergent integral a value equal to its finite part.

Finite part integral

In this post we will consider divergent integrals arising from singularities at the end point of integration, in particular, divergent integrals of the form
\begin{equation}
\int_0^c \frac{f(x)}{x^{j+\nu}} \mathrm{d}x, j=1, 2, 3, \dots, 0 \leq \nu <1, 0 < c \leq \infty .
\end{equation} We assume that $f(x)$ is integrable in any finite interval of the real line. When it happens that the upper limit of integration is infinite we mean the divergent integral
\begin{equation}
\int_0^{\infty} \frac{f(x)}{x^{j+\nu}} \, \mathrm{d}x=\lim_{c\rightarrow\infty} \int_0^c \frac{f(x)}{x^{j+\nu}} \, \mathrm{d}x
\end{equation}So essentially it is sufficient to consider the case for finite $c$ and then obtain the limit $c\rightarrow\infty$ as needed.

The finite part of the divergent integral is obtained as follows. Let $0 < \epsilon < c$. Then we have the convergent integral
\begin{equation}
\int_{\epsilon}^{c} \frac{f(x)}{x^{j+\nu}}\mathrm{d}x
\end{equation}The divergent integral is recovered in the limit $\epsilon\rightarrow 0$. For non-zero $\epsilon$ we can divide the above integral in two terms,
\begin{equation}
\int_{\epsilon}^{c} \frac{f(x)}{x^{j+\nu}}\mathrm{d}x = C_{\epsilon}+D_{\epsilon},
\end{equation} where $C_{\epsilon}$ is the group of terms that have a finite limit as $\epsilon\rightarrow 0$ and $D_{\epsilon}$ is the group of terms that diverge in the same limit. The finite part of the divergent integral is simply obtained by dropping the diverging term $D_{\epsilon}$ and leaving only the finite limit of $C_{\epsilon}$. That is
\begin{equation}
\mathrm{FPI}\!\!\int_0^c \frac{f(x)}{x^{j+\nu}} \mathrm{d}x = \lim_{\epsilon\rightarrow 0} C_{\epsilon} .
\end{equation}

Finite part integration of the given integral

Let us now attempt to evaluate the given integral above using the finite part of the divergent integrals that appear in the distribution of the integration in the two terms of the integrand. We then calculate the finite part of the divergent integral $\int_0^{\infty}e^{-a x}x^{-3/2}\mathrm{d}x$. Let $0 < \epsilon < c <\infty$. Then the integral
\begin{equation}
\int_{\epsilon}^{c} \frac{e^{-a x}}{x^{3/2}} \mathrm{d}x
\end{equation}converges. We will eventually take the limit $c\rightarrow\infty$, in accordance with the fact that the integral is an improper integral. We expand the exponential and perform term by term integration, which is allowed as the infinite series has an infinite radius of convergence. We obtain
\begin{equation}\label{integ}
\int_{\epsilon}^{c} \frac{e^{-a x}}{x^{3/2}}\mathrm{d}x = -\frac{2}{\sqrt{c}} + \frac{2}{\sqrt{\epsilon}} + \sum_{k=1}^{\infty} \frac{(-1)^k a^k}{k!(k-1/2)} \left(c^{k-1/2}-\epsilon^{k-1/2}\right).
\end{equation}We identify the divergent and convergent terms as
\begin{equation}
D_{\epsilon}= \frac{2}{\sqrt{\epsilon}}
\end{equation}\begin{equation}
C_{\epsilon} = -\frac{2}{\sqrt{c}} + \sum_{k=1}^{\infty} \frac{(-1)^k a^k}{k!(k-1/2)} \left(c^{k-1/2}-\epsilon^{k-1/2}\right)
\end{equation} Clearly $D_{\epsilon}$ becomes infinite in the limit $\epsilon\rightarrow 0$; this the source of the divergence of the integral $\int_0^{\infty}e^{-a x}x^{-3/2}\mathrm{d}x$. On the other hand $C_{\epsilon}$ has a well defined (finite) limit as $\epsilon\rightarrow 0$.

The finite part of the divergent integral $\int_0^{c} e^{-a x} x^{-3/2}$ is the part of equation \eqref{integ} that remains finite as $\epsilon\rightarrow$, that is by dropping $D_{\epsilon}$ and keeping only $C_{\epsilon}$ in the limit. The finite part is then given by
\begin{equation}
\mathrm{FPI}\!\!\int_{0}^{c} \frac{e^{-a x}}{x^{3/2}} \mathrm{d}x = -\frac{2}{\sqrt{c}} + 2 \frac{1-e^{-a c}}{\sqrt{c}} – 2 \sqrt{\pi a}\, \mathrm{erf}(\sqrt{ac})
\end{equation} where we have performed the indicated summation and $\mathrm{erf}(z)$ is the error function.

Finally the desired finite part integral is obtained in the limit $c\rightarrow\infty$, and is given by
\begin{equation}
\mathrm{FPI}\!\!\int_{0}^{\infty} \frac{e^{-a x}}{x^{3/2}} \mathrm{d}x = -2\sqrt{\pi a},
\end{equation} which follows from the property of the error function that $\mathrm{erf}(\infty)=1$. Using this result we obtain the difference
\begin{equation}
\mathrm{FPI}\!\!\int_0^{\infty} \frac{e^{-a x}}{x^{3/2}} \mathrm{d}x – \mathrm{FPI}\!\!\int_0^{\infty} \frac{e^{-b x}}{x^{3/2}} \mathrm{d}x = 2 (\sqrt{b}-\sqrt{a}) \sqrt{\pi}.
\end{equation} which correctly reproduces the value of the given convergent integral.

We have just demonstrated that the given integral can be distributed provided we interpret the arising divergent integrals as finite part integrals, that is,
\begin{equation}
\int_0^{\infty} \frac{e^{-a x} – e^{-b x}}{x^{3/2}} \mathrm{d}x =\mathrm{FPI}\!\!\int_0^{\infty} \frac{e^{-a x}}{x^{3/2}} \mathrm{d}x – \mathrm{FPI}\!\!\int_0^{\infty} \frac{e^{-b x}}{x^{3/2}} \mathrm{d}x .
\end{equation} Does that mean that we can arbitrarily perform term by term evaluation provided we use the finite part when divergent integrals arise in the process?

A counter example

No. Let us give an example how naive term by term integration with divergent integrals may lead to missing terms. Let us consider the integral
\begin{equation}\label{integral}
\int_0^a \frac{\mathrm{d}x}{\lambda + x}=-\ln\lambda + \ln(\lambda + a), \;\;\; \lambda, a>0 .
\end{equation}Let us binomially expand $(\lambda+ x)^{-1}$ as follows
\begin{equation}
\frac{1}{\lambda + x} = \sum_{j=0}^{\infty} (-1)^j \frac{\lambda^j}{x^{j+1}},
\end{equation} and then substitute the expansion back in the left hand side of equation \eqref{integral}, followed by term by term integration. The whole process yields the formal infinite series
\begin{equation}\label{infinite}
\int_0^a \frac{\mathrm{d}x}{\lambda + x} \leadsto \sum_{j=0}^{\infty} (-1)^j \lambda^j \int_0^a \frac{1}{x^{j+1}} \mathrm{d}x
.
\end{equation} Notice that the integrals are divergent, which requires us that we decide on how to assign values to them.

Again we interpret them as finite part integrals (FPI). The finite part of the divergent integrals are obtained as follows. For a given integer $j=0,1,2,\dots$, let $\epsilon>0$ and consider the integral
\begin{equation}\label{divergent}
\int_{\epsilon}^{a} \frac{\mathrm{d}x}{x^{j+1}} = \left\{\begin{array}{ll}
\ln a – \ln \epsilon &, \;\; j=0 \\
-\frac{1}{j}\left(\frac{1}{a^j}- \frac{1}{\epsilon^{j}}\right) &, \;\; j=1,2,\dots
\end{array} \right. .
\end{equation} The finite part of the divergent integral $\int_0^ax^{-j-1} \mathrm{d}x$ is the term in equation \eqref{divergent} that has a finite limit as $\epsilon\rightarrow 0$. In other words, the finite part is obtained by just simply dropping all terms in equation \eqref{divergent} that diverge in the said limit. We then obtain
\begin{equation}
\mathrm{FPI}\int_0^a \frac{\mathrm{d}x}{x^{j+1}} =\left\{ \begin{array}{ll}
\ln a &,\;\; j=0 \\
-\frac{1}{j a^j} &, \;\;j=1, 2, \dots
\end{array}
\right.
\end{equation}

We substitute these finite part integrals back into the formal infinite series \eqref{infinite} to yield
\begin{equation}\label{formal}
\ln a + \sum_{s=1}^{\infty} \frac{(-1)^{s+1}}{s} \frac{\lambda^s}{a^s}
\end{equation} Now we expand the right hand side of equation \eqref{integral} for sufficiently large $a$, and obtain
\begin{equation}\label{exact}
\int_0^a \frac{\mathrm{d}x}{\lambda+ x}=-\ln\lambda + \ln a + \sum_{s=1}^{\infty} \frac{(-1)^{s+1}}{s} \frac{\lambda^s}{a^s} .
\end{equation} Comparing the two expansions \eqref{formal} and \eqref{exact}, we find them to agree except on the dominant term in the second expansion.

This demonstrates how naive term by term integration and careless interpretation of divergent integral can lead to missing terms.

How to do finite part integration

What is happening? How come we have exact equality in the first instance of our application of the finite part of a divergent integral but we have non-equality in the second instance? The paper The problem of missing terms in term by term integration involving divergent integrals addresses this question and shows how we can make the finite part integral work by lifting the integration in the complex plane.

$The contour of integration $\mathrm{C}$. The branch cut of $\log z$ and $z^{-\nu}$ are indicated by the zigzag line, which is the positive real axis.$

The contour of integration $\mathrm{C}$. The branch cuts of $\log z$ and $z^{-\nu}$ are indicated by the zigzag line.

The key to the proper use of the finite integral is its representation as a contour integral in the complex plane. We summarize below our main results. In the following, the branch cuts of $\log z$ and $z^{-\nu}$ for $0<\nu<1$ are chosen along the positive real axis. Also the complex extension of a real-valued function $f(x)$ is the complex valued function obtained from from $f(x)$ by replacing $x$ with the complex number $z$, e.g., the complex extension of $e^x$ is $e^z$.

The following Theorems give the complex contour integral representation of the finite part of the divergent integrals relevant to us here.

Theorem 1. Let the complex extension, $f(z)$, of $f(x)$ be analytic in some neighborhood of the interval $[0,a]$. If $f(0)\neq 0$, then
\begin{equation}\label{result1}
\mathrm{FPI}\!\!\int_0^a\frac{f(x)}{x^{n+1}}\mathrm{d}x=\frac{1}{2\pi i}\int_{\mathrm{C}} \frac{f(z)}{z^{n+1}} \left(\log z-\pi i\right)\mathrm{d}z,
\end{equation}for $n=0, 1, \dots$, where the contour $\mathrm{C}$ is the contour straddling the branch cut of $\log z$ starting from $a$ and ending at $a$ itself, as depicted in the Figure.

Theorem 2. Let the complex extension, $f(z)$, of $f(x)$ be analytic in some neighborhood of the interval $[0,a]$. If $f(0)\neq 0$, then
\begin{equation}\label{branch}
\mathrm{FPI}\!\!\int_0^a \frac{f(x)}{x^{m+\nu}}\mathrm{d}x = \frac{1}{\left(\mathrm{e}^{-2\pi \nu i } -1\right)} \int_{\mathrm{C}} \frac{f(z)}{z^{m+\nu}} \mathrm{d}z,
\end{equation}for $m=1, 2, \dots$ and $0<\nu<1$, where the contour $\mathrm{C}$ is the contour straddling the branch cut of $z^{-\nu}$ starting from $a$ and ending at $a$ itself, as depicted in the Figure.

To make use of the contour integral representation of the finite part integral, the given integral must itself be cast in an integral representation consistent with the integral representation of the finite part integral. To this end we have the following results:

Proposition 3. Let $g(x)$ be integrable in the interval $[0,a]$ and that its complex extension $g(z)$ is analytic in a region containing the interval $[0,a]$ in its interior and holomorphic elsewhere. Then
\begin{equation}\label{contourep1}
\int_0^a g(x) \, \mathrm{d}x = \frac{1}{2\pi i} \int_{\mathrm{C}} g(z) \log z \, \mathrm{d}z – \sum_k \mathrm{Res}\left[\log z \, g(z)\right]_{z_k},
\end{equation}where the contour $\mathrm{C}$ is the same contour in Theorem-1, and the $z_k$’s are the poles of $g(z)$ enclosed by $\mathrm{C}$, with no pole of $g(z)$ lying along $\mathrm{C}$.

Proposition 4. Let $g(x)$ be integrable in the interval $[0,a]$ and that its complex extension $g(z)$ is analytic in a region containing the interval $[0,a]$ in its interior. Then
\begin{eqnarray}
\int_0^a x^{-\nu} g(x) \mathrm{d}x &=& \frac{1}{\left(\mathrm{e}^{-2\pi \nu i}-1\right)} \int_{\mathrm{C}} z^{-\nu} g(z)\, \mathrm{d}z \nonumber \\
&&- \frac{2\pi i}{\left(\mathrm{e}^{-2\pi \nu i}- 1\right)} \sum_k \mathrm{Res}\left[z^{-\nu} g(z)\right]_{z=z_k},
\end{eqnarray} where $\mathrm{C}$ is the same contour in Theorem-2, and the $z_k$’s are the poles of $g(z)$ enclosed by $\mathrm{C}$, with no pole lying on the contour $\mathrm{C}$.

Going back to the examples

We now apply above results for our first integral. Theorem-2, together with Proposition-2, applies. Our divergent integral corresponds to $m=1$, $\nu=1/2$, and $f(x)=(e^{-a x}-e^{-b x})$. The complex extension of $f(x)$ is $f(z)=(e^{-a z}-e^{-b z})$ which is an entire function, i.e., it has no pole. Then the integral assumes the contour integral form
\begin{equation}
\int_0^{\infty} \frac{e^{-a x} – e^{-b x}}{x^{3/2}} \mathrm{d}x = -\frac{1}{2} \int_{\mathrm{C}} \frac{e^{-a z} – e^{-b z}}{z^{3/2}} \mathrm{d}z . \label{contour}
\end{equation}
Here is the key. Notice that the integration in the right hand side is an integral along a contour that does not pass through the origin. This allows us to distribute the integration through the two terms giving us. We recognize that each term is just the finite part of the divergent integral so that we have
\begin{equation}
\int_0^{\infty} \frac{e^{-a x} – e^{-b x}}{x^{3/2}} \mathrm{d}x = \mathrm{FPI}\!\!\int_0^{\infty} \frac{e^{-a x}}{x^{3/2}} \mathrm{d}x
– \mathrm{FPI}\!\!\int_0^{\infty} \frac{e^{-b x}}{x^{3/2}} \mathrm{d}x
\end{equation}This establishes rigorously the use of the finite part integral of the divergent integrals in evaluating the original integral.

Now let us resolve the missing term in the term by term integration in the second example. Again to employ finite part integration, we lift the integration in the complex plane. Here Theorem-1, together with Proposition-1, applies. First we have to to represent the given integral as a contour integral in the complex plane. The function involved is the function $(\lambda + z)^{-1}$ which has simple pole at $z=-\lambda$. Here the contour $\mathrm{C}$ has to enclose the pole. Then we have the contour integral representation
\begin{equation}
\int_0^a \frac{1}{\lambda + x} \mathrm{d}x = – \mathrm{Res}\left[\log z \cdot \frac{1}{\lambda + z}\right]_{z=-\lambda} + \frac{1}{2\pi i} \int_{\mathrm{C}}\frac{1}{\lambda + z} \, \log z \, \mathrm{d}z
\end{equation} We make the trivial change in the second integral by $\log z = (\log z – i \pi) + i\pi$. The reason for this is that we should be able to eventually recognize the finite part integral in the ensuing expansion. Then
\begin{eqnarray}
\int_0^a \frac{1}{\lambda + x} \mathrm{d}x &=& – \log(-\lambda) + \frac{1}{2\pi i} \int_{\mathrm{C}}\frac{1}{\lambda + z} \, (\log z – i \pi) \, \mathrm{d}z \nonumber \\
&&\hspace{38mm}+ \frac{1}{2\pi i} \int_{\mathrm{C}}\frac{i\pi}{\lambda + z} \, \mathrm{d}z
\end{eqnarray}

With $\log(-\lambda)=\log(\lambda e^{i \pi})= \ln\lambda + i\pi$ and $\int_{\mathrm{C}} (\lambda+z)^{-1} \mathrm{d}z = 2\pi i$, we obtain the contour integral representation of the original integral as
\begin{equation}
\int_0^a \frac{1}{\lambda + x} \mathrm{d}x = – \ln\lambda + \frac{1}{2\pi i} \int_{\mathrm{C}}\frac{1}{\lambda + z} \, (\log z – i \pi) \, \mathrm{d}z
\end{equation}We insert the expansion
\begin{equation}
\frac{1}{\lambda + z} = \sum_{j=0}^{\infty} (-1)^j \frac{\lambda^j}{z^{j+1}},
\end{equation} back into the second term and interchange the order of integration and summation. We obtain
\begin{equation}
\int_0^a \frac{1}{\lambda + x} \mathrm{d}x = – \ln\lambda + \sum_{j=0}^{\infty}(-1)^j \lambda^j\frac{1}{2\pi i} \int_{\mathrm{C}} \frac{1}{z^{j+1}}(\log z – i \pi) \, \mathrm{d}z
\end{equation}The interchange is justified provided the condition $\lambda<a$ is satisfied. Now we recognize that the contour integrals in the series are just the finite part integrals. Then
\begin{equation}
\int_0^a \frac{1}{\lambda + x} \mathrm{d}x = – \ln\lambda + \sum_{j=0}^{\infty}(-1)^j\lambda^j\; \mathrm{FPI}\!\!\int_0^a \frac{1}{x^{j+1}}\mathrm{d}x
\end{equation}This completely reproduces the correct expansion obtained above. Observe that the missing term comes from the simple pole of the integrand.