Derivation of an Integral Expression of the Euler-Mascheroni Constant

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = -\gamma
\label{eq:1608111}
\tag{1}
\end{equation}

ii boros mollThere are of course an enormous number of integral expressions of the Euler-Mascheroni Constant. This particular expression appeared on the cover of Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals by George Boros and Victor Moll and was derived in section 9.3 of the book. Here, I will show this derivation and fill in the steps as well as parts that were not shown in the book.

Let us begin by providing a basic definition of the Euler-Mascheroni Constant
\begin{equation}
\gamma := \lim_{n \to \infty} H_{n} – \mathrm{ln}(n)
\label{eq:1608112}
\tag{2}
\end{equation}
where \(H_{n}\) are the harmonic numbers defined as
\begin{equation}
H_{n} = \displaystyle\sum_{k=1}^{n} \frac{1}{k}
\label{eq:1608113}
\tag{3}
\end{equation}

We will break up the derivation into parts to make it easier to follow.

1. Exercise 9.3.1 – Prove
\begin{equation}
\int\limits_{0}^{1} \frac{1-(1-x)^{n}}{x} \mathrm{d} x = H_{n}
\label{eq:1608114}
\tag{4}
\end{equation}

\begin{align}
\label{eq:1608115}
\tag{5}
\int\limits_{0}^{1} \frac{1-(1-x)^{n}}{x} \mathrm{d} x & = \int\limits_{0}^{1} \frac{1-y^{n}}{1-y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1}{1-y} \mathrm{d} y \, – \int\limits_{0}^{1} \frac{y^{n}}{1-y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \displaystyle\sum_{k=0}^{\infty} y^{k} \mathrm{d} y \,\, – \int\limits_{0}^{1} y^{n} \displaystyle\sum_{k=0}^{\infty} y^{k} \mathrm{d} y \\
& = \displaystyle\sum_{k=0}^{\infty} \frac{y^{k+1}}{k+1} \big|_{0}^{1} \,\, – \displaystyle\sum_{k=0}^{\infty} \frac{y^{k+n+1}}{k+n+1} \big|_{0}^{1} \\
& = (1 + \frac{1}{2} + \frac{1}{3} + \dots) \, – (\frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} + \dots) \\
& = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} = H_{n}
\end{align}

Notes:

  • \(|x| < 1\)
  • Let \(y=1-x\)
  • Expand \(1/(1-y)\) as a geometric series \begin{equation} \frac{1}{1-y} = \displaystyle\sum_{k=0}^{\infty} y^{k} \end{equation}
  • The integrals are improper, we have dropped the limits for clarity.
  • We assumed that the integrals converged uniformly so that we could integrate term by term. We acknowledge that this should be proven, but the assumption is warranted based on the form of the integrand and experience.

2. Proposition 9.3.1 – The Euler constant is given by
\begin{equation}
\gamma = \int\limits_{0}^{1} \frac{1-\mathrm{e}^{-x}}{x} \mathrm{d} x \,\, – \int\limits_{1}^{\infty} \frac{\mathrm{e}^{-x}}{x} \mathrm{d} x
\label{eq:1608116}
\tag{6}
\end{equation}
The proof in the book is fairly straightforward but there are some missing steps.

Make the substitution \(x = \frac{y}{n}\) in \eqref{eq:1608114} to obtain
\begin{align}
H_{n} & = \int\limits_{0}^{n} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \int\limits_{1}^{n} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \int\limits_{1}^{n} \frac{1}{y} \mathrm{d} y \,\, – \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \mathrm{ln}(n) \, – \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d} y
\label{eq:1608117}
\tag{7}
\end{align}

Applying \(\lim n \to \infty\) to equation \eqref{eq:1608117}, noting the limit definition of
\begin{equation}
\mathrm{e}^{\pm x} = \lim_{n \to \infty} \left(1 \pm \frac{x}{n} \right)^{n}
\end{equation}
and invoking equation \eqref{eq:1608112} yields our result.

3. Proposition 9.3.2 – Euler’s constant is given by equation \eqref{eq:1608111}.

Here we require one obvious alteration of the integral in light of \eqref{eq:1608117} and a non obvious trick.

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = \int\limits_{0}^{1} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x + \int\limits_{1}^{\infty} \mathrm{e}^{-x} \mathrm{ln}(x) \mathrm{d} x = \mathrm{I}_{1} + \mathrm{I}_{2}
\label{eq:1608118}
\tag{8}
\end{equation}

Integrating \(\mathrm{I}_{2}\) by parts and applying the appropriate limit yields
\begin{equation}
\mathrm{I}_{2} = \int\limits_{1}^{\infty} \frac{\mathrm{e}^{-x}}{x} \mathrm{d} x
\label{eq:1608119}
\tag{9}
\end{equation}

For \(\mathrm{I}_{1}\) use the substitution
\begin{equation}
\mathrm{e}^{-x} = -\frac{\mathrm d}{\mathrm d x} (\mathrm{e}^{-x} – 1),
\end{equation}
integrate by parts and apply the appropriate limit to obtain
\begin{equation}
\mathrm{I}_{1} = -\int\limits_{0}^{1} \frac{\mathrm{e}^{-x} – 1}{x} \mathrm{d} x
\label{eq:16081110}
\tag{10}
\end{equation}
Substituting equations \eqref{eq:16081110} and \eqref{eq:1608119} into \eqref{eq:1608118} and using \eqref{eq:1608116} yields our final result.

A Clever Integration Trick

This trick is from Integration for Engineers and Scientists by William Squire via The Handbook of Integration by Daniel Zwillinger.

Noting that
\begin{equation}
\frac{1}{x} = \int\limits_{0}^{\infty} \mathrm{e}^{-xt} \mathrm{d} t
\end{equation}
we can replace \(\frac{1}{x}\) in an integrand with its integral expression and reverse the order of integration to simplify evaluation of the integrals. Of course the integral must converge uniformly to allow us to reverse the order of integration and we must have \(t>0\).

Let us use this trick to evaluate
\begin{equation}
\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x} \mathrm{d} x
\end{equation}
for \(a,b > 0\).
Using our trick yields
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x} \mathrm{d} x & = \int\limits_{0}^{\infty}\int\limits_{0}^{\infty} \mathrm{e}^{-xt} (\mathrm{e}^{-ax}-\mathrm{e}^{-bx}) \mathrm{d} t \mathrm{d} x \\
& = \int\limits_{0}^{\infty}\int\limits_{0}^{\infty} \mathrm{e}^{-(a+t)x}-\mathrm{e}^{-(b+t)x} \mathrm{d} x \mathrm{d} t \\
& = -\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-(a+t)x}}{a+t} – \frac{\mathrm{e}^{-(b+t)x}}{b+t} |_{0}^{\infty} \mathrm{d} t \\
& = \int\limits_{0}^{\infty} \frac{1}{a+t} – \frac{1}{b+t} \mathrm{d} t \\
& = \lim_{R \to \infty} \mathrm{ln}\frac{a+t}{b+t} |_{0}^{R} \\
& = \mathrm{ln}\left(\frac{b}{a}\right)
\end{align}

The conventional way to handle this integral is to recognize that it is the Frullani integral
\begin{equation}
\int\limits_{0}^{\infty} \frac{f(ax) – f(bx)}{x} \mathrm{d} x = [f(\infty) – f(0)]\mathrm{ln}\left(\frac{a}{b}\right)
\end{equation}
A simple substitution yields our result.

For the Frullani integral, we must have the existence of \(f(\infty)\) and \(f(0)\) using the appropriate limits. For other conditions on the integral as well as proofs, see

  1. On Cauchy-Frullani Integrals by A. M. Ostrowski. Use DOI 10.1007/BF02568143 with Sci-Hub to access the paper.
  2. On the Theorem of Frullani by Juan Arias-De-Reyna. Use DOI 10.2307/2048376 with Sci-Hub to access the paper.

Integrate \(\int_{0}^{\infty} \frac{\mathrm{ln}(x^{2}+a^{2})}{x^{2}+b^{2}} \mathrm{d} x\)

For \(a,b > 0\),

\begin{equation}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}(x^{2}+a^{2})}{x^{2}+b^{2}} \mathrm{d} x = \frac{\pi}{b} \mathrm{ln}(a+b)
\label{eq:160806a1}
\tag{1}
\end{equation}

appeared on page 52 of Rediscovery of Malmsten’s integrals, their evaluation by contour integration methods and some related results by Iaroslav V. Blagouchine. This is a fascinating paper with many interesting results. In future blog posts, I will present some of Blagouchine’s results and solve some of the exercise problems that he proposed. For now, I will do this integral mainly to highlight a common trick used to evaluate contour integrals with logarithms of binomials.

The trick is to begin with a different integrand
\begin{equation}
f(z) = \frac{\mathrm{ln}(z+ia)}{z^{2}+b^{2}} = \frac{\mathrm{ln}(z+ia)}{(z-ib)(z+ib)}
\label{eq:160806a2}
\tag{2}
\end{equation}

Using the following contour

semi circle contour with pole on im axis

 

we note that a first order pole at \(z=ib\) is inside of the contour so we have
\begin{equation}
Res_{z=ib}[f(z)] = \frac{\mathrm{ln}(ib+ia)}{i2b} = \frac{\mathrm{ln}(i)+\mathrm{ln}(a+b)}{i2b} = \frac{i\frac{\pi}{2}+\mathrm{ln}(a+b)}{i2b}
\label{eq:160806a3}
\tag{3}
\end{equation}

\begin{align}
\oint\limits_{C} f(z) \mathrm{d} z & = i2\pi Res_{z=ib}[f(z)] = \frac{i\pi^{2}}{2b} + \frac{\pi}{b}\mathrm{ln}(a+b) \\
& = \lim_{R \to \infty} \int\limits_{-R}^{R} f(x) \mathrm{d} x + \int\limits_{C_{1}} f(z) \mathrm{d} z
\label{eq:160806a4}
\tag{4}
\end{align}

The second integral goes to 0 via the ML estimate. The first integral will be broken in half and we use the substitution \(y=-x\) to obtain
\begin{equation}
\int\limits_{-\infty}^{0} \frac{\mathrm{ln}(x+ia)}{x^{2}+b^{2}} \mathrm{d} x = \int\limits_{0}^{\infty} \frac{\mathrm{ln}(-y+ia)}{y^{2}+b^{2}} \mathrm{d} y
\label{eq:160806a5}
\tag{5}
\end{equation}

Adding the two halves of the integral together, we have the following in the numerator
\begin{equation}
\mathrm{ln}(-x+ia) + \mathrm{ln}(x+ia) = i\pi + \mathrm{ln}(x-ia) + \mathrm{ln}(x+ia) = \mathrm{ln}(x^{2}+a^{2})
\end{equation}

Now we have
\begin{equation}
\oint\limits_{C} f(z) \mathrm{d} z = \int\limits_{0}^{\infty} \frac{\mathrm{ln}(x^{2}+a^{2})}{x^{2}+b^{2}} \mathrm{d} x +
i\pi \int\limits_{0}^{\infty} \frac{1}{x^{2}+b^{2}} \mathrm{d} x
\label{eq:160806a6}
\tag{6}
\end{equation}

Equating real and imaginary parts of equations \eqref{eq:160806a6} and \eqref{eq:160806a4} yields our original result plus a bonus integral
\begin{equation}\int\limits_{0}^{\infty} \frac{1}{x^{2}+b^{2}} \mathrm{d} x = \frac{\pi}{2b}
\end{equation}
which we could have obtained via the inverse tangent function.

Note that the trick allowed the limits of the integral to work out with the semi circular contour and we recovered the original integrand. This is a standard trick but surprisingly I have read some complex analysis texts that do not cover it.

A Laplace Transform Proof of \(\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\)

This proof appeared in Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals by George Boros and Victor Moll. Here, I fill in the steps of the proof.

We begin with 3 fundamental results of the Laplace transform: the basic definition, convolution, and the Laplace transform of a convolution.

\begin{equation}
\mathcal{L}[f(t)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-st} f(t) \mathrm{d} t
\label{eq:1608061}
\tag{1}
\end{equation}

\begin{equation}
(f*g)(t) = \int\limits_{0}^{t} f(\tau) g(t-\tau) \mathrm{d} \tau
\label{eq:1608062}
\tag{2}
\end{equation}

\begin{equation}
\mathcal{L}[f*g] = \mathcal{L}[f]\mathcal{L}[g]
\label{eq:1608063}
\tag{3}
\end{equation}

We begin with the following two functions
\begin{equation}
f(t) = t^{x-1} \quad \mathrm{and} \quad g(t) = t^{y-1}
\label{eq:1608064}
\tag{4}
\end{equation}

These functions were chosen for two reasons. First, they appear in the basic integral definition of the beta function
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{1} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:1608065}
\tag{5}
\end{equation}
and, as we now show, their Laplace transforms result in gamma functions
\begin{align}
\mathcal{L}[f(t)] & = \mathcal{L}[t^{x-1}] = \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x-1} \mathrm{d} t \\
& = \frac{1}{s^{x}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x-1} \mathrm{d} v \\
& = \frac{\Gamma(x)}{s^{x}}
\label{eq:1608066}
\tag{6}
\end{align}
where we have used the substitution \(v = st\). Likewise, taking the Laplace transform of \(g(t)\) yields,
\begin{equation}
\mathcal{L}[g(t)] = \mathcal{L}[t^{y-1}] = \frac{\Gamma(y)}{s^{y}}
\label{eq:1608067}
\tag{7}
\end{equation}

Now we substitute equations \eqref{eq:1608067}, \eqref{eq:1608066}, and \eqref{eq:1608062} into \eqref{eq:1608063}
\begin{equation}
\frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} = \mathcal{L}\left[\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau \right]
\label{eq:1608068}
\tag{8}
\end{equation}

To evaluate the integral, we let \(\tau = tu\)
\begin{align}
\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau & = t^{x+y-1} \int\limits_{0}^{1} u^{x-1} (1 – u)^{y-1} \mathrm{d} u \\
& = t^{x+y-1} \mathrm{B}(x,y)
\label{eq:1608069}
\tag{9}
\end{align}

Thus, we have
\begin{align}
\mathcal{L}[t^{x+y-1} \mathrm{B}(x,y)] & = \frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} \\
& = \mathrm{B}(x,y) \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x+y-1} \mathrm{d} t \\
& = \mathrm{B}(x,y) \frac{1}{s^{x+y}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x+y-1} \mathrm{d} v \\
& = \mathrm{B}(x,y) \frac{\Gamma(x+y)}{s^{x+y}}
\label{eq:16080610}
\tag{10}
\end{align}

Combining the first and last results from the right hand side of the above equation yields our final result
\begin{equation}
\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
\end{equation}

A Derivation of the Beta Function Representation \(\mathrm{B}(x,y) = 2^{1-x-y} \int_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t\)

I used the beta function representation

\begin{equation}
\mathrm{B}(x,y) = 2^{1-x-y} \int\limits_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t
\label{eq:1608051}
\tag{1}
\end{equation}

here to evaluate an integral. Now I will derive this result.

We begin with the basic integral definition of the beta function

\begin{equation}
\mathrm{B}(x,y) = \int\limits^{1}_{0} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:1608052}
\tag{2}
\end{equation}

then use the substitution \(t = \frac{z-a}{b-a}\), this yields

\begin{align}
\mathrm{B}(x,y) & = \frac{1}{b-a} \int\limits^{b}_{a} \left(\frac{z-a}{b-a}\right)^{x-1} \left(\frac{b-z}{b-a}\right)^{y-1} \mathrm{d} z \\
& = (b-a)^{-x-y+1} \int\limits^{b}_{a} (z-a)^{x-1} (b-z)^{y-1} \mathrm{d} z \\
& = 2^{-x-y+1} \int\limits_{-1}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z \\
& = 2^{-x-y+1} \left[ \int\limits_{-1}^{0} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z + \int\limits_{0}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z \right]
\label{eq:1608053}
\tag{3}
\end{align}

Note that we let \(a=-1\) and \(b=1\).

For the rightmost integral, we let \(z=-w\) so that

\begin{equation}
\int\limits_{0}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z = \int\limits_{0}^{1} (1-w)^{x-1} (1+w)^{y-1} \mathrm{d} w
\label{eq:1608054}
\tag{4}
\end{equation}

Substituting equation \eqref{eq:1608054} into equation \eqref{eq:1608053} yields equation \eqref{eq:1608051}.

The original substitution, \(t = \frac{z-a}{b-a}\) is a useful method of transforming the limits of integration from 0 to 1 into a to b.

Integrate \(\int_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x\)

From Victor Moll’s attempt to solve all of the integrals in Gradshteyn and Ryzhik, we have

\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2}
\label{eq:1608041}
\tag{1}
\end{equation}

Moll’s solution is at the link, so let us solve this another way. We first use the substitution used by Moll, let \(x=av\) to obtain

\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \int\limits_{0}^{1} (1-v^{2})^{n-1/2} \mathrm{d} v
\label{eq:1608042}
\tag{2}
\end{equation}

We designate the integral on the right hand side as I and make the substitution \(z = v^{2}\)

\begin{equation}
\mathrm{I} = \int\limits_{0}^{1} (1-v^{2})^{n-1/2} \mathrm{d} v = \frac{1}{2} \int\limits_{0}^{1} z^{-1/2} (1-z)^{n-1/2} \mathrm{d} z
\label{eq:1608043}
\tag{3}
\end{equation}

From the beginning this integral looked like a gamma or beta function, now, in this form it is obviously a beta function. We then have

\begin{equation}
\mathrm{I} = \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\label{eq:1608044}
\tag{4}
\end{equation}

and thus

\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\label{eq:1608045}
\tag{5}
\end{equation}

Now we must show that the right hand sides of equations \eqref{eq:1608041} and \eqref{eq:1608045} are equal or

\begin{equation}
\mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right) \overset{\underset{\mathrm{?}}{}}{=} \pi \frac{(2n-1)!!}{(2n)!!}
\label{eq:1608046}
\tag{6}
\end{equation}

We begin by expressing the beta function in terms of gamma functions

\begin{equation}
\mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right) = \frac{\Gamma(\frac{1}{2}) \Gamma(n + \frac{1}{2})}{\Gamma(n+1)} = \sqrt{\pi}
\frac{\Gamma(n + \frac{1}{2})}{\Gamma(n+1)}
\label{eq:1608047}
\tag{7}
\end{equation}

Now we invoke the following two relationships between the gamma function and the double factorial, found at Wikipedia and Wolfram Math World respectively

\begin{equation}
(2n-1)!! = \frac{2^{n} \Gamma(n + \frac{1}{2})}{\sqrt{\pi}} \quad \mathrm{and} \quad (2n)!! = 2^{n}n! = 2^{n} \Gamma(n+1)
\end{equation}

Substituting these expressions into \eqref{eq:1608046} shows that it is indeed an equality.

Putting everything together, we have
\begin{equation}
\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2} = a^{2n} \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\end{equation}

Notes on Richard Dedekind’s Was sind und was sollen die Zahlen? by David E. Joyce

Richard Dedekind (1831–1916) published in 1888 a paper entitled Was sind und was sollen die Zahlen? variously translated as What are numbers and what should they be? or, as Beman did, The Nature of Meaning of Numbers. Dedekind added a second preface to the second edition in 1893. The second edition was translated into English by Wooster Woodruff Beman in 1901.

Much of Dedekind’s terminology and notation as translated by Beman differs from the modern standards, so I offer these notes, using current terminology and notation, to make his work more understandable.

The pdf file is here.

Even today, the importance of rigorously establishing the nature and properties of real numbers is underappreciated. In a book I cite in a previous postTheory and Application of Infinite Series, the author, Knopp, takes great pains to emphasize the properties of real numbers. A reviewer on Amazon.com complained about this, but this is the very attitude that Knopp seeks to dispel.

Shifting the Expansion Point of a Power Series

If we have a function that has been expanded in a power series about the point \(x = x_{0}\)

\begin{equation}
f(x) = \sum\limits_{n=0}^{\infty} a_{n} (x – x_{0})^{n}
\end{equation}

and if the radius of convergence of this series is non-zero, then we can generate a power series expansion about a new point \(x = x_{1}\) if this point is in the radius of convergence of the original power series. The new series is [1]

\begin{equation}
f(x) = \sum\limits_{k=0}^{\infty} b_{k} (x – x_{1})^{k}
\end{equation}

\begin{equation}
b_{k} = \sum\limits_{n=0}^{\infty}
\begin{pmatrix}
n+k \\
k
\end{pmatrix}
a_{n+k} (x_{1} – x_{0})^{n}
\end{equation}

While this is clearly not practical for paper and pencil solutions, it does appear that it could be useful for computer applications. If you had an algorithm for determining the coefficients of the original power series and you wanted to generate many new power series with different expansion points, then this formulation would allow one to do this by only have to store the coefficients of the original power series. One can imagine having a series of expansion points, \(x_{1}s\), running a loop over them, and using this algorithm to generate different representations of the given function as power series.

[1] Theory and Application of Infinite Series by Konrad Knopp.

Integrate \(\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x\)

This integral appeared in Inside Interesting Integrals by Paul Nahin in the problem set of chapter 3. Using Wolfram Alpha, we get

\begin{equation}
\int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = \pi
\label{eq:1}
\tag{1}
\end{equation}

Nahin suggests the following trig substitution, \(x = \cos(2y)\).

While the form of the integrand certainly does suggest that some type of trig substitution will work, let us do it with another method. If we write the integral as

\begin{equation}
\int\limits_{-1}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x
\end{equation}

this looks like a beta function. From Higher Transcendental Functions (Bateman Manuscript), Volume 1, Section 1.5.1, equation 10, we see

\begin{equation}
\mathrm{B}(x,y) = 2^{1-x-y} \int\limits_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t
\label{eq:2}
\tag{2}
\end{equation}

Let us begin with the original integral and the right half of the interval of integration

\begin{equation}
\int\limits_{0}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x = \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:3}
\tag{3}
\end{equation}

Now, let us consider

\begin{equation}
\int\limits_{0}^{1} (1+x)^{-\frac{1}{2}}(1-x)^{\frac{1}{2}} \mathrm{d} x = \int\limits_{0}^{1}\sqrt{\frac{1-x}{1+x}} \mathrm{d} x
\label{eq:4}
\tag{4}
\end{equation}

We let \(x=-y\) to obtain

\begin{equation}
-\int\limits_{0}^{-1} \sqrt{\frac{1+y}{1-y}} \mathrm{d} y,
\label{eq:5}
\tag{5}
\end{equation}

which we can rewrite as

\begin{equation}
\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:6}
\tag{6}
\end{equation}

Adding the right hand side of equation \eqref{eq:3} and equation \eqref{eq:6} yields our original integral

\begin{equation}
\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x + \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = \int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:7}
\tag{7}
\end{equation}

Likewise, adding the left hand sides of equations \eqref{eq:4} and \eqref{eq:3} yields

\begin{equation}
\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x + \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x =
\int\limits_{0}^{1} (1+x)^{-\frac{1}{2}}(1-x)^{\frac{1}{2}} \mathrm{d} x + \int\limits_{0}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x
\end{equation}

If we combine this result into one integral and rearrange the integrand, we see that it is the same as the integral in \eqref{eq:2} with

\begin{equation}
x=\frac{3}{2} \,\, \mathrm{and} \,\, y=\frac{1}{2}
\end{equation}

Putting it all together, we have

\begin{equation}
\int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = 2\mathrm{B}\left(\frac{3}{2},\frac{1}{2}\right) = \pi
\end{equation}

Integrate \(\int^{\infty}_{0}\frac{e^{-px^{2}} – e^{-qx^{2}}}{x^{2}} \mathrm{d}x\)

This integral appeared in Paul Nahin’s very interesting book Inside Interesting Integrals. Nahin begins with a completely different integral and derives this one. Let us evaluate the integral directly and then redo it with Nahin’s method.

We begin by breaking up the integral and looking at each piece. So we have

\begin{equation}
\mathrm I = \int\limits^{\infty}_{0} x^{-2}\mathrm{e}^{-px^{2}} \mathrm{d}x.
\end{equation}

This looks very similar to a definition of the gamma function:

\begin{equation}
\Gamma(z) = \int\limits^{\infty}_{0} x^{z-1}\mathrm{e}^{-x} \mathrm{d}x.
\end{equation}

We make the substitution \(y = px^{2}\)

\begin{equation}
\mathrm I = \frac{\sqrt{p}}{2} \int\limits^{\infty}_{0} \mathrm{e}^{-y} y^{-\frac{3}{2}} \mathrm{d}y.
\end{equation}

Invoking the gamma function yields

\begin{equation}
\mathrm I = \frac{\sqrt{p}}{2} \Gamma\Big(-\frac{1}{2}\Big) = -\sqrt{p}\sqrt{\pi}.
\end{equation}

Treating the other part of the original integral involving \(q\) yields our final result

\begin{equation}
\int\limits^{\infty}_{0}\frac{\mathrm{e}^{-px^{2}} – \mathrm{e}^{-qx^{2}}}{x^{2}} \mathrm{d}x = \sqrt{\pi}(\sqrt{q}-\sqrt{p}).
\end{equation}

As I mentioned earlier, Nahin derived this result beginning with an entirely different integral. A casual glance at the original integral should make us suspect that this is the case as it is clear that both parts of the integrand are identical. In other words, why solve the original integral as opposed to the integral that I used at the beginning of the analysis. Such is the case with many of the results in Inside Interesting Integrals. This is the result of working backward, yielding an evaluated integral via some methods as opposed to starting from an integral that one wants to evaluate. I am not criticizing this approach, as it has resulted in an enormous number of useful integral evaluations. Indeed, it can create an unlimited number of evaluated integrals. Also, such “accidental” integrals can result from contour integration even when directly attacking a given integral. Consider that it often happens that upon the last step in evaluating an integral via contour integration, one equates real and imaginary parts in which one is the solution to the original integral while the other is a bonus.

Let us now see how Nahin achieved his result. He begins with

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x^{2}} \mathrm{d}x
\end{equation}

for which Nahin derived the answer of \(\frac{1}{2} \sqrt{\pi} \) earlier in the book. What is interesting here is that this integral can be done easily with the gamma function by letting \( x^{2} = y \). This quickly results in

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-x^{2}} \mathrm{d}x = \frac{1}{2} \int\limits_{0}^{\infty} \mathrm{e}^{-y} y^{-1/2} \mathrm{d} y =
\frac{1}{2} \Gamma\Big(\frac{1}{2}\Big) = \frac{1}{2} \sqrt{\pi}.
\end{equation}

If someone saw this, then they would immediately recognize that the integral sought can be evaluated via the gamma function as I did above. Nevertheless, let us continue with Nahin’s analysis.

Nahin makes a change of variable, \(x = t\sqrt{a} \) to introduce the parameter \(a\), and thus obtains

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-at^{2}} \mathrm{d}t = \frac{1}{2}\frac{\sqrt{\pi}}{\sqrt{a}}
\end{equation}

Then he invokes a useful and interesting trick. He integrates the equation with respect to \(a\), between two arbitrary end points, and changes the order of integration. Changing the order of integration requires some care, as it is only valid if the integral converges uniformly. Here, the integral is just a gamma function, which we know converges uniformly. This is usually the case for “well behaved”, “non-crazy” integrals. So, Nahin has for the left hand side

\begin{equation}
\int\limits_{p}^{q}\left\{\int\limits_{0}^{\infty} \mathrm{e}^{-at^{2}} \mathrm{d}t\right\}\mathrm{d}a = \int\limits_{0}^{\infty}\left\{\int\limits_{p}^{q}\mathrm{e}^{-at^{2}} \mathrm{d}a\right\} \mathrm{d}t = \int\limits_{0}^{\infty}\frac{\mathrm{e}^{-pt^{2}} – \mathrm{e}^{-qt^{2}}}{t^{2}} \mathrm{d}t.
\end{equation}

The right hand side yields

\begin{equation}
\int\limits_{p}^{q}\frac{1}{2}\frac{\sqrt{\pi}}{\sqrt{a}} \mathrm{d}a = \sqrt{\pi}(\sqrt{q}-\sqrt{p}).
\end{equation}

And thus we have our result.