improper integrals - uneigentliche Integrale
The idea here is to include integrals that have an infinite interval of integration and integrands that become infinite within the interval of integration.
integrals over infinite intervals - Integral über einem unbeschränktem Interval
example 1
The graph of ${f(x)= \frac{1}{x^2}}$
, the line ${x=1}$
, the x-axis and the line ${x=k}$
enclose an area.
a) Find the area ${A(k)}$
in terms of k.
b) Investigate the limit of the area ${A(k)}$
as ${k \rightarrow \infty}$
solution
a) the area is the integral of f(x) between x=1 and x=k. So, we find the anti-derivative of f(x). $${A(k) = \int \limits_1^{+ \infty} \frac{1}{x^2}\mathrm{d}x = F(k) - F(1)}$$
b) investigate the behaviour as ${k \rightarrow \infty}$
$${\int \limits_1^{+ \infty} \frac{1}{x^2}\mathrm{d}x= \lim_{k \to \infty} \int \limits_a^{k}{ \frac{1}{x^2}\mathrm{d}x} = \lim \limits_{k \to \infty} \left[ -\frac{1}{x}\right]_1^k = \lim \limits_{k \to \infty}{\left( 1 - \frac{1}{k}\right)} = 1}$$
example 2
$${\int \limits_1^{+ \infty} \frac{1}{x}\mathrm{d}x= \lim_{k \to \infty} \int \limits_a^{k}{ \frac{1}{x}\mathrm{d}x} = \lim \limits_{k \to \infty} \left[ \ln{x}\right]_1^k = \lim \limits_{k \to \infty}{\ln{k}} = + \infty}$$
The limit diverges. So, the improper integral does not exits.
definition
The improper integral (uneigentliche Integral) of f over the interval
${[a, + \infty)}$
is defined as $${\int \limits_a^{+ \infty} f(x) \mathrm{d}x = \lim_{k \to \infty} \int \limits_a^{k}{ f(x) \mathrm{d}x}}$$ in the case where the limits exists, the improper integral is said to converge, and the value is calculated by evaluating the integral. If the limit does not exist, the improper integral is said to diverge.
example 3
$${\int \limits_{-2}^{-\frac{1}{2}} \frac{1}{x}\mathrm{d}x = \left[ \ln{x}\right]_{-2}^{-\frac{1}{2}} = {\ln{(-\frac{1}{2})} - \ln{(-2)}}}$$
But, the natural logarithm is only defined for x > 0.
However, we can use the symmetry of the function and then the integral does have a meaning.
$${\int \limits_{-2}^{-\frac{1}{2}} \frac{1}{x}\mathrm{d}x = - \int \limits_{\frac{1}{2}}^{2} \frac{1}{x}\mathrm{d}x = \left[-\ln{x}\right]_{\frac{1}{2}}^{2} = {-(\ln{2} - \ln{\frac{1}{2}})} = {\ln{\frac{1}{2}} - \ln{2}}}$$
exercises - Übungen
Berechnen Sie, sofern sie existieren, die folgenden uneigentlichen Integrale
a) ${\int \limits_1^{ \infty} \frac{1}{x^3}\mathrm{d}x \quad }$
b) ${\int \limits_2^{ \infty} 8x^{-5}\mathrm{d}x \quad}$
c) ${\int \limits_1^{ \infty} \frac{1}{\sqrt{x}}\mathrm{d}x \quad}$
d) ${\int \limits_{-\infty}^{-2} \frac{x + 1}{x^4}\mathrm{d}x}$
vertical asymptotes 
At $x = 0$
neither $\frac{1}{x}$
nor $\ln (0)$
are defined.
Therefore, it is not possible to evaluate any integral where the limits have opposite signs like
$${\int \limits_{-1}^{ 2} \frac{1}{x}\mathrm{d}x}$$ because it has two infinitely large areas.
So, in general $${\int \limits_{a}^{ b} \frac{1}{x}\mathrm{d}x = \bigg[\ln {\left|{x}\right|}\bigg]_a^b}$$
if $a, b \neq 0$
and either $a, b > 0$
or $a, b < 0$
(i.e., a and b have the same sign)
A function can only be integrated if it is continuous (stetig) over the interval of integration.
integral in the form ${\int \frac{f^\prime(x)}{f(x)}\mathrm{d}x}$
example 4
Use the substitution $u = 3x^2 + 1$
to evaluate
$${\int \frac{6x}{3x^2 + 1} \mathrm{d}x}$$
An integral in the form ${\int \frac{f^\prime(x)}{f(x)}\mathrm{d}x}$
can be converted in to the form ${\int \frac{1}{u}\mathrm{d}u}$
by substituting $u = f(x)$
If the graph of f(x) is continuous (between a and b) then $${\int \frac{f^\prime(x)}{f(x)}\mathrm{d}x = \ln {\bigg|{f(x)}\bigg|} + C }$$ or $${\int\limits_{a}^{ b} \frac{f^\prime(x)}{f(x)}\mathrm{d}x = \bigg[ \ln {\bigg|{f(x)}\bigg|}\bigg]_a^b }$$
exercises - Übungen
Find
a) ${\int \frac{3}{2x - 5}\mathrm{d}x \quad }$
b) ${\int \frac{x}{1 + x^2}\mathrm{d}x \quad}$
c) ${\int \tan{x} \mathrm{d}x}$
Note - you might have to rewrite the integrals slightly.