Log Integral Varieties

{Log Integral Varieties}

{abstract}

We discuss computation of indefinite integrals of the form $$I_{m,n}:=int x^{m}(ln x)^{n} dx, (m,n)inmathbb{N}^{2},$$

for a range of values $(m,n)$, frequently encountered in an advanced level calculus course. Attention is drawn to reduction formula for definite integrals of $I_{m,n}$ over the unit interval $[0,1]$. Several interesting integral formulae in terms of said parameters are derived. A working knowledge of integration by parts, and recurrence relations will be required.

{Introduction}

The integral at the focus of our enquiry has vast applications in analysis. In particular, it provides a means for calculating integrals of functions which have no elementary form, and affords one to interpolate the factorial function $n!$, beyond the standard definition for non-negative integers: $n!=ncdot(n-1)cdot(n-2)ldots1$, that is

egin{equation}

n!=int_{0}^{1}(-ln x)^{n}dx, n otin{-1,-2,-3,ldots}

end{equation}

which affords surprising calculations, for example one finds $(frac{1}{2})!=frac{sqrt{pi}}{2}$. The treatment of the integral $I_{m,n}$ discussed here, requires little more than the emph{integration by parts formula}, employed in the calculation of integrals of the form $int f(x)g(x)dx$. As with most calculations of this nature, we must first prioritise how best to assign the functions $(f,g)$ to the components

$frac{dv}{dx}, u(x)$

appearing in said formula, which may be written

egin{equation}

int ufrac{dv}{dx}dx=uv-int vfrac{du}{dx}dx

end{equation}

where care must be taken over our choice of $frac{dv}{dx}, u(x)$. For poor designation would further complicate the integral $int vfrac{du}{dx}dx$, often yielding a problem far more taxing than initially prescribed. Drawing our attention to the integral at hand

$$I_{m,n}:=int x^{m}(ln x)^{n} dx, (m,n)inmathbb{N}^{2},$$

one should always opt to assign the monomial, $x^{m}$, and logarithm, $(ln x)^{n}$, as follows

$$frac{dv}{dx}=x^{m}, u=(ln x)^{n}$$

as our motivation here will be to reduce the index on the logarithm throughout our calculations. This being achieved by virtue of the $frac{du}{dx}$ term appearing in the far integral of (2). For problems of this flavour, met by most students, one should find $ninmathbb{N}leqslant2$ and $minmathbb{Z}geqslant-1$, ensuring we iterate emph{by parts} at most twice, with this being governed by the index of the logarithm.

We now present some calculations relevant to an A level course, culminating in several reduction formulae encountered in further mathematics courses leading to a proof of (1).

{Case $(m,n)=(0,1)$}

We commence with calculation of the integral

$$I_{0,1}=intln x dx$$

one should have sound knowledge of this proof. Moreover, recall of this result will lessen the number of iterations required for the parts formula when computing for higher $n$. With $u(x)=ln x$ and $frac{dv}{dx}=1$ application of (2) yields

$$intln x dx=intln xcdot 1 dx=xln x-int xleft(frac{1}{x} ight)dx=xln x-x+c$$

{Case $(m,n)=(-1,n)$}

We now consider the following integral

$$I_{-1,n}=intfrac{1}{x}left(ln x ight)^{n}dx, n eq-1$$

which slightly varies our mode of approach. This is an example of the standard pattern of the chain rule

$$int f`(x)left(f(x) ight)^{n}dx=frac{1}{n+1}(f(x))^{n+1}+c$$

hence, we have

$$intfrac{1}{x}left(ln x ight)^{n}dx=frac{1}{n+1}(ln x)^{n+1}+c$$

for the degenerate case, $n=-1$, we again draw on a standard result from integral calculus, namely the logarithmic form

$$intfrac{f`(x)}{f(x)}dx=ln|f(x)|+c$$

for $(m,n)=(-1,-1)$, we observe

$$I_{-1,-1}=intfrac{1}{xln x}dx=intfrac{left(frac{1}{x} ight)}{ln x}dx=ln|ln x|+c$$

{Case $(m,n)=(m,2)$}

This next example demonstrates multiple use of (2)

$$I_{m,2}=int x^{m}(ln x)^{2}dx$$

with $frac{dv}{dx}=x^{m}, u=(ln x)^{2}$ we have, after one iteration

$$int x^{m}(ln x)^{2}dx=frac{x^{m+1}(ln x)^{2}}{m+1}-frac{1}{m+1}int x^{m+1}cdotfrac{2}{x}(ln x)dx$$

one should observe that further application of (2), be required for the far integral

$$frac{1}{m+1}int x^{m+1}cdotfrac{2}{x}(ln x)dx=frac{2}{(m+1)^{2}}x^{m+1}(ln x)-frac{2}{(m+1)^{2}}int x^{m+1}cdotfrac{1}{x}dx$$

$$=frac{2}{(m+1)^{2}}x^{m+1}(ln x)-frac{2}{(m+1)^{3}}x^{m+1}+c$$

combining the results, we have

$$I_{m,2}=frac{x^{m+1}(ln x)^{2}}{m+1}-frac{2}{(m+1)^{2}}x^{m+1}(ln x)+frac{2}{(m+1)^{3}}x^{m+1}+c$$

noting the degenerate case $m=0$

$$I_{0,2}=int (ln x)^{2}dx=x(ln x)^{2}-2x(ln x)+2x+c$$

which is useful in emph{volume of revolution} calculations $piint y^{2}dx$

{Further reduction formulae}

We conclude our analysis with several reduction formulae, useful in a multitude of science and engineering applications.

Here we define $I^{*}_{m,n}$ as the definite integral

$$I^{*}_{m,n}:=int_{0}^{1}x^{m}(ln x)^{n}dx, (m,n)inmathbb{N}^{2}$$

application of (2) yields

$$I^{*}_{m,n}=left[frac{x^{m+1}(ln x)^{n}}{m+1} ight]_{0}^{1}-frac{n}{m+1}int_{0}^{1}x^{m}(ln x)^{n-1}dx$$

one may question evaluation of the parenthesis term with $x=0$. However, it can be shown for any given $ninmathbb{N}$, that

$$lim_{x ightarrow0^{+}}x(ln x)^{n}=0$$

resulting in the recurrence relation

$$I^{*}_{m,n}=-frac{n}{m+1}I^{*}_{m,n-1}$$

we proceed with a further $r-1$ iterations, leading to

$$I^{*}_{m,n}=left(-frac{n}{m+1} ight)left(-frac{n-1}{m+1} ight)left(-frac{n-2}{m+1} ight)ldotsleft(-frac{n-r+1}{m+1} ight)I^{*}_{m,n-r}$$

$$=frac{(-1)^{r}}{(m+1)^{r}}frac{n!}{(n-r)!}I^{*}_{m,n-r}$$

given $r=n$, we have the reduction formula

$$I^{*}_{m,n}=frac{(-1)^{n}n!}{(m+1)^{n}}I^{*}_{m,0}, ext{with} I^{*}_{m,0}=int_{0}^{1}x^{m}dx=frac{1}{m+1}$$

arriving at the integral result

$$I^{*}_{m,n}=int_{0}^{1} x^{m}(ln x)^{n} dx=frac{(-1)^{n}n!}{(m+1)^{n+1}}$$

which may also be written as

$$int_{0}^{1} x^{m}(-ln x)^{n} dx=frac{n!}{(m+1)^{n+1}}$$

with $m=0$, we arrive the factorial result (1).

{Addendum}

Interestingly, if $m=n$

egin{equation}int_{0}^{1} x^{n}(-ln x)^{n} dx=frac{n!}{(n+1)^{n+1}}

end{equation}

and considering the known series expansion

$$frac{1}{x^{x}}=sum_{n=0}^{infty}frac{x^{n}(-ln x)^{n}}{n!}$$

by drawing on (3), one may observe the rather pleasing result

$$int_{0}^{1}frac{1}{x^{x}}dx=sum_{n=0}^{infty}frac{1}{n!}int_{0}^{1}x^{n}(-ln x)^{n}dx=sum_{n=0}^{infty}frac{1}{(n+1)^{n+1}}$$

which may be written as

$$int_{0}^{1}frac{1}{x^{x}}dx=sum_{n=1}^{infty}frac{1}{n^{n}}$$

an integral result commonly referred to as emph{Sophomore`s Dream}.