$e^{-x^2}$ - 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

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.

Example 1 ((inspection, substitution))

(a) $\int (2x+3)/(x^2+3x-2)\mathrm{dx}$
(b) ${\int }_0^{1}1/((4-x)(x-1{)}^{1/2})\mathrm{dx}$

(a) $\int e^{-x}\cos (x)\mathrm{dx}$
(b) $\int x^2{e}^x\mathrm{dx}$

$\int (\cos (x))/(3+\cos (x))\mathrm{dx}$ (difficult)

Example 4 ((inspection, by parts))

${\int }_0^x{e}^{-x^2}/(x^2+(1/2{))}^2\mathrm{dx}$

(Very difficult). Your answer should include ${\int }_0^x{e}^{-x^2}\mathrm{dx}$ left as it is. This can not be evaluated for general $x$ in terms of elementary functions.

Example 5

${\int }_{-\infty }^{\infty }{e}^{-x^2}\mathrm{dx}$

(Impossible or not !!!)

Let $I={\int }_{-\infty }^{\infty }{e}^{-x^2}\mathrm{dx}$

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

$I={\int }_{-\infty }^{\infty }{e}^{-z^2}\mathrm{dz}$

Multiplying the two expressions for $I$ we obtain

$I^2={\int }_{-\infty }^{\infty }{e}^{-x^2}\mathrm{dx}{\int }_{-\infty }^{\infty }{e}^{-z^2}\mathrm{dz}$

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-x^2}{e}^{-z^2}\mathrm{dx}\mathrm{dz}$

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-(x^2+z^2)}\mathrm{dx}\mathrm{dz}$

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 $\theta$, i.e.

$x^2+z^2=r^2$

and

$\mathrm{dx}\mathrm{dz}=r\mathrm{dr}d\theta$

and complete the evaluation. If all goes well you should find that $I={\pi }^{1/2}$

Example 6

${\int }_0^{\infty }{x}^3/(e^x-1)\mathrm{dx}$

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

$\sum _{n=1}^{\infty }1/n^4={\pi }^4/90$

Let

$I={\int }_0^{\infty }{x}^3/(e^x-1)\mathrm{dx}$

and noting that since $e^{-x}\le 1$, then throughout the range of integration we can Taylor expand,and so

$x^3/(e^x-1)=e^{-x}{x}^3(1/(1-e^{-x}))(1+e^{-x}+e^{-2x}+...)$

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

$I={\pi }^4/15$

Example 6

Without carrying out the integration, use your answer in the previous example to show that

${\int }_0^{\infty }{x}^3/(e^x+1)\mathrm{dx}=7/8.{\pi }^4/15$

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