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Integrals with natural logarithm...

chisigma

Well-known member
Feb 13, 2012
1,704
Recently in the 'Challenge Forum' the following integral has been proposed...

$$\int_{0}^{\infty} \frac{\ln x}{x^{2}+ a^{2}}\ d x\ (1)$$

Scope of this note is to illustrate a general procedure to engage integrals like (1) in elementary way, i.e. without use comnplex analysis tecniques. The preliminary is the evaluation of the following indefinite integral, which doesn't appear in most 'Integration Manuals'...


$$\int x^{m}\ \ln^{n} x\ d x\ (2)$$

... where m and n are non negative integers. Proceeding with standard integration by part we first obtain...

$$\int x^{m}\ \ln^{n} x\ d x = \frac{x^{m+1}}{m+1}\ \ln^{n} x - \frac{n}{m+1}\ \int x^{m}\ \ln^{n-1} x\ d x\ (3)$$

... so that the integral in second term is the original integral with exponent of the logarithm is n-1 instead of n. Repeating n times this procedure we arrive to the final result...


$$\int x^{m}\ \ln^{n} x\ d x = x^{m+1}\ \sum_{i=0}^{n} (-1)^{i}\ \frac{n!}{(n-i)!\ (m+1)^{i+1}}\ \ln^{n-i} x + c\ (4)$$

The (4) is itself important and it will be the basis of all successive steps....


Comments or question about this thread can be submitted in a specific thread open in the 'Commentary Forum'...

http://mathhelpboards.com/commentary-threads-53/commentary-integrals-natural-logarithm-5287.html

Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
The indefinite integral we found in previous post…

$$\int x^{m}\ \ln^{n} x\ d x = x^{m+1}\ \sum_{i=0}^{n} (-1)^{i}\ \frac{n!}{(n-i)!\ (m+1)^{i+1}}\ \ln^{n-i} x + c\ (1)$$

… now will be used in the particular case…

$$\int_{0}^{1} x^{m}\ \ln^{n} x\ d x = |x^{m+1}\ \sum_{i=0}^{n} (-1)^{i}\ \frac{n!}{(n-i)!\ (m+1)^{i+1}}\ \ln^{n-i} x|_{0}^{1} = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}}\ (2)$$

... and the (2) can be used the more general integral...


$$\int_{0}^{1} f(x)\ \ln^{n} x\ dx\ (3)$$

... where f(*) is an analytic function in the interval 0 < x < 1, i.e. is...

$$f(x) = \sum_{k=0}^{\infty} a_{k}\ x^{k}\ (4)$$

... so that we can write...

$$\int_{0}^{1} f(x)\ \ln^{n} x\ dx = \sum_{k=0}^{\infty} a_{k}\ \int_{0}^{1} x^{k}\ \ln^{n} x\ dx = (-1)^{n} n!\ \sum_{k=0}^{\infty} \frac{a_{k}}{(k+1)^{n+1}}\ (5) $$

One of the most simple cases is $f(x)= \frac{1}{1-x}$ for which is $a_{k}=1\ \forall k$, so that is...


$$\int_{0}^{1} \frac{\ln^{n} x}{1-x}\ dx = (-1)^{n} n!\ \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+1}}\ = (-1)^{n}\ n!\ \zeta(n + 1)\ (6) $$

... where $\zeta(*)$ is the so called 'Riemann Zeta Function'. In next post we will examine other interesting similar integrals...

Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Writing in different form the conclusion of the previous post we can say that, given an analytic function...

$\displaystyle f(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n}\ (1)$

... then the following relation holds...

$\displaystyle \sum_{n=1}^{\infty} \frac{a_{n-1}}{n^{k}} = \frac{(-1)^{k-1}}{(k-1)!}\ \int_{0}^{1} f(x)\ \ln^{k-1} x\ dx\ (2)$

The first term of (2) belongs to a family of series called Dirichlet Series and they play an important role in Analytic Number Theory. The most famous of them is the so called 'Riemann Zeta Function' where $\displaystyle f(x) = \frac{1}{1-x}$ so that $\forall n$ is $a_{n}=1$ and we have...

$\displaystyle \zeta (k) = \sum_{n=1}^{\infty} \frac{1}{n^{k}} = \frac{(-1)^{k-1}}{(k-1)!}\ \int_{0}^{1} \frac{ \ln^{k-1} x}{1-x}\ dx\ (3)$

Also very important is the Dirichlet Series which derives from the so called 'Moebious Function', a discrete function defined as...

$\displaystyle \mu(n) = \begin {cases} (-1)^{\omega(n)} & \text{if}\ \omega(n)=\Omega(n)\\ 0 & \text{if}\ \omega(n)<\Omega(n)\end{cases}\ (4)$

... where $\omega(n)$ is the number of distinct primes dividing the number $n$ and $\Omega(n)$ is the number of prime factors of $n$, counted with multiplicities; clearly $\omega(n) \le \Omega(n)$. This Dirichlet series is...

$\displaystyle \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{k}} = \frac{1}{\zeta(k)}\ (5)$

Now if we want to find for it an expression like (2) we introduce the function...

$\displaystyle \rho(x) = \sum_{n=0}^{\infty} \mu(n+1)\ x^{n}\ (6)$

... that is represented in the figure...




... so that we arrive to the expression...

$\displaystyle \frac{1}{\zeta(k)} = \frac{(-1)^{k-1}}{(k-1)!}\ \int_{0}^{1} \rho(x)\ \ln^{k-1} x\ dx\ (7)$

The same procedure can be extended to other Dirichlet series...

Kind regards

$\chi$ $\sigma$
 
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