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Python 3
python
python3
#%config InlineBackend.figure_format = 'svg'
from pylab import *
Many times, a change of variable can improve the regularity of the
integrand.
:::{prf:example} Singular integrand
$$
I = \int_0^b \frac{f(x)}{\sqrt{x}} \ud x, \qquad \textrm{$f$ is smooth}
$$
With
\begin{gather}
x = u^2, \qquad 0 \le u \le \sqrt{b} \
\Downarrow \
I = 2 \int_0^{\sqrt{b}} f(u^2) \ud u
\end{gather}
:::
+++
:::{prf:example} Singular derivative
$$
\int_0^1 \sin(x) \sqrt{1-x^2} \ud x = 2\int_0^1 u^2 \sqrt{2-u^2} \sin(1-u^2) \ud u
$$
by change of variables $u = \sqrt{1-x}$ . The first form has singular
derivative at $x=1$ while the second form has infinitely smooth
integrand.
:::
+++
:::{prf:example} Infinite interval of integration
$$
I = \int_1^\infty \frac{f(x)}{x^p} \ud x, \qquad p > 1
$$
Assume $\lim_{x \to \infty} f(x)$ exists and $f(x)$ is smooth on $[1,\infty)$ . Change of variable
\begin{gather}
x = \frac{1}{u^\alpha}, \qquad \ud x = -\frac{\alpha}{u^{1+\alpha}} \ud u, \quad \alpha > 0 \
\Downarrow \
I = \alpha \int_0^1 u^{(p-1)\alpha-1} f(1/u^\alpha) \ud u
\end{gather}
Integrand can be singular at $u=0$ , so pick $\alpha$ so that the exponent $(p-1)\alpha-1$ is large $(\ge 1)$ .
As a specific example consider
\begin{gather}
I = \int_1^\infty \frac{f(x)}{x\sqrt{x}} \ud x \
x = \frac{1}{u^4} \quad\implies\quad I = 4 \int_0^1 u f(1/u^4) \ud u
\end{gather}
For $x \to \infty$ , $f(x)$ behaves like
\begin{gather}
f(x) = c_0 + \frac{c_1}{x} + \frac{c_2}{x^2} + \ldots \
\Downarrow \
uf(1/u^4) = c_0 u + c_1 u^5 + c_2 u^9 + \ldots
\end{gather}
and the integrand behaves nicely as $u \to 0$ .
:::
+++
:::{prf:example} Chebyshev quadrature
$$
I = \int_{-1}^1 \frac{f(x)}{\sqrt{1-x^2}} \ud x
$$
For this weight function, the orthogonal polynomials are Chebyshev polynomials. The integration nodes are roots of $T_n(x) = \cos(n \cos^{-1}x)$ (Chebyshev points of I kind)
$$
x_j = \cos\left( \frac{2j-1}{2n}\pi \right), \qquad j=1,2,\ldots,n
$$
and weights
$$
w_j = \frac{\pi}{n}
$$
so that
$$
I \approx I_n = \frac{\pi}{n} \sum_{j=1}^n f(x_j)
$$
This corresponds to doing the change of variable $x = \cos\theta$ and applying the mid-point rule in the $\theta$ variable
$$
I = \int_0^\pi f(\cos\theta) \ud\theta \approx \frac{\pi}{n} \sum_{j=1}^n f(\cos\theta_j), \qquad \theta_j = \frac{(2j-1)\pi}{2n}
$$
:::
Analytic treatment of singularity Consider an integral with a singular term
$$
I = \int_0^b f(x) \log(x) \ud x
$$
Since $\log(x)$ is singular at $x=0$ , we have to compute this with some care. Divide the interval into two parts
$$
I = \int_0^\epsilon f(x) \log(x) \ud x + \int_{\epsilon}^b f(x) \log(x) \ud x =: I_1 + I_2
$$
$I_2$ can be computed by some standard technique and $I_1$ is computed semi-analytically. Assume $f(x)$ has convergent Taylor series on $[0,\epsilon]$
$$
f(x) = \sum_{j=0}^\infty a_j x^j, \qquad x \in [0,\epsilon]
$$
Then
$$
I_1 = \int_0^\epsilon \left( \sum_j a_j x^j \right) \log(x) \ud x = \sum_{j=0}^\infty \frac{a_j \epsilon^{j+1}}{j+1} \left[ \log(\epsilon) - \frac{1}{j+1} \right]
$$
+++
:::{prf:example} Singular integrand
$$
I = \int_0^{4\pi} \cos(x) \log(x) \ud x
$$
Then
$$
I_1
&= \int_0^{\epsilon} \cos(x) \log(x) \ud x \\
&= \epsilon[\log(\epsilon)-1] - \frac{\epsilon^3}{6}[\log(\epsilon) - 1/3] + \frac{\epsilon^5}{600}[\log(\epsilon) - 1/5] + \ldots
$$
The terms in the above series become smaller, so we can truncate this
series at some term.
:::
Consider an integral with a singular term
$$
I(f) = \int_a^b w(x) f(x) \ud x
$$
where $w(x)$ is singular and $f(x)$ is a nice function. Let ${ f_n }$ be a sequence of approximations to $f$ such that
$|w|$ is integrable.
$\lim_{n\to\infty} \norm{f - f_n}_\infty = 0$
The integrals
$$
I_n(f) = \int_a^b w(x) f_n(x) \ud x
$$
can be analytically computed.
Then
$$
|I(f) - I_n(f)| \le \norm{f-f_n}_\infty \int_a^b |w(x)| \ud x \to 0
$$
+++
:::{prf:example} Singular integrand
$$
I(f) = \int_0^b f(x) \log(x) \ud x
$$
Let $f_n$ be piecewise linear approximation on a uniform partition
$$
x_j = jh, \quad j=0,1,\ldots,n \qquad\textrm{where}\qquad h = \frac{b}{n}
$$
and
$$
f_n(x) = \frac{1}{h}[ (x_j - x) f_{j-1} + (x-x_{j-1}) f_j], \qquad x_{j-1} \le x \le x_j
$$
for which we know the error estimate from
$$
\norm{f-f_n}_\infty \le \frac{h^2}{8} \norm{f''}_\infty
$$
Then, with $I_n(f) = I(f_n)$
$$
|I(f) - I_n(f)| \le \frac{h^2}{8} \norm{f''}_\infty \int_0^b |\log(x)| \ud x \to 0
$$
$I_n(f)$ can be computed analytically. In particular, in the first sub-interval $[x_0,x_1]=[0,h]$
$$
f_n(x) = \frac{h-x}{h} f_0 + \frac{x}{h} f_1
$$
we have
$$
\int_{x_0}^{x_1} \log(x) f_n(x) \ud x = \frac{f_0}{h} \int_0^h (h-x)\log(x) \ud x + \frac{f_1}{h} \int_0^h x \log(x) \ud x
$$
and these are finite since
$$
\int_0^h \log(x) \ud x = h[\log(h)-1], \qquad \int_0^h x\log(x)\ud x = \frac{h^2}{4}[ 2\log(h) - 1 ]
$$
The integrals in other intervals can also be computed analytically.
:::