Hard integrals, and strategies for
integration
By Rupert on June 3, 1998 :
Hiya please send me the hardest integral that you can think
of
thanx
By Gordon Lee (gtyl2) on June 3,
1998 :
Rupert,
I think that whether an integral is hard is very subjective and
that most of the time. If you are trying to integrate an
expression because you need it to solve your problem, most of the
time, you won't be able to do the integrating analytically! There
is a branch of Mathematics call Numerical Analysis and it deals
with computing often non-analytical (ie. you can't integrate!)
problems numerically!
I will give you an example of something you can't
integrate:
sin(x)/x
I will probably send you a few more later.
Gordon
By Gordon Lee (gtyl2) on June 4,
1998 :
As promised more problematic integrals:
e-x2 - the exp part of the pgf of the normal distribution
1/(cos x-c)1/2 - arises in calculating the time period of a pendulum
Gordon
By Demetris Andreou Demetriou (dad20) on
June 12, 1998 :
Health Warning: The following discussion (and examples) are
appropriate for people at the end of their first year in A-Level Math or
equivalent. The difficulty of a specific integral is very ambiguous and depends on the
techniques one has at hand, as well as whether the object is simply to carry
out the integration, or carry out the integration using a certain technique.
To the extend of the material covered in A-level math the object is to carry
out the integral. There are many elementary tricks/techniques that allow one
to tackle integrals that at first sight might seem impossible to do. These
include
- inspection,
- substitutions,
- partial fractions,
- integration by parts,
- use of standard results, e.g. another integral or series.
The following examples illustrate some of the above techniques.
Example 1 ((inspection, substitution))
(a) ò(2x+3)/(x2+3x-2)dx
(b) ò01 1/((4-x)(x-1)1/2) dx
Example 2 ((by parts))
(a) òe-x cos(x) dx
(b) òx2 ex dx
Example 3 ((substitution, partial fractions))
ò(cos(x))/(3+cos(x))dx (difficult)
Example 4 ((inspection, by parts))
ò0x e-x2/(x2+(1/2))2 dx
(Very difficult). Your answer should include ò0x e-x2 dx left as
it is. This can not be evaluated for general x in terms of elementary
functions.
Example 5
ò-¥¥ e-x2 dx
(Impossible or not !!!)
Let I=ò-¥¥ e-x2 dx
To evaluate it we are going to employ in a clever waysome of the properties
of the exponential function. One can equally write the above in terms of a
different variableof integration
I=ò-¥¥e-z2 dz
Multiplying the two expressions for I we obtain
I2=ò-¥¥ e-x2 dxò-¥¥ e-z2 dz
=ò-¥¥ ò-¥¥ e-x2e-z2 dx dz
=ò-¥¥ ò-¥¥ e-(x2+z2) dx dz
which is a double integral extending over the x z plane. Consider then
expressing the integration over this plane in terms of polar coordinates, r
and q, i.e.
x2+z2=r2
and
dx dz=r dr dq
and complete the evaluation. If all goes well you should find that
I=p1/2
Example 6
ò0¥ x3/(ex-1)dx
Only for people doing further Maths, or special papers)
The following is not the most elegant way of evaluating the above integral,
BUT it does not require anything more than knowledge of SERIES EXPANSIONS, and
the standard result, (assumed to be given) that
Let
I=ò0¥ x3/(ex-1) dx
and noting that since e-x £ 1, then throughout the range of integration
we can Taylor expand,and so
x3/(ex-1)=e-xx3(1/(1-e-x))(1+e-x+e-2x+...)
Can you prove the above? Consider expanding in a series (1-e-x)-1.
What happens if you move e-x into the brackets? Can you see how the
summation result quoted above can be of any use? If all goes well you should
find that
I=p4/15
Example 6
Without carrying out the integration, use your answer in the previous example
to show that
ò0¥ x3/(ex+1) dx=7/8.p4/15
_____________________________
Demetris Andreou Demetriou