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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 wether 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)ò0 11/((4-x)(x-1)½)dx


Example 2 ((by parts))

a) òe-xcos(x)dx , b) òx2exdx


Example 3 ((substitution, partial fractions))

ò(cos(x))/(3+cos(x))dx (difficult)


Example 4 ((inspection,by parts ))

ò0 x e-x2/(x2+½)2 dx
(Very difficult). Your answer should include ò0 x e-x2dx 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 way some of the properties of the exponential function. One can equally write the above in terms of a different variable of integration

I = ò-¥ ¥ e-z2 dz

Multiplying the two expressions for I we obtain
I2 = ò-¥ ¥ e-x2 dx ò-¥ ¥ e-z2 dz
= ò-¥ ¥ ò-¥ ¥ e-x2 e-z2 dx dz
= ò-¥ ¥ ò-¥ ¥ e-(x2+z2) dx dz
which is a double integral extending over the xz plane. Consider then expressing the integration over this plane in terms of polar coordinates, r and q, i.e.
x2+z2=r2

and
dxdz = rdrdq

and complete the evaluation. If all goes well you should find that
|2{I = p½}

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

S¥ n=1 1/n4 = p4/90

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

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Demetris Andreou Demetriou