# Math Help - Help with Integration.

1. ## Help with Integration.

Hi all. It may be simple, but I do not seem to recall it. How can I solve the following

As an example u may like to set n to any integer, say, 5.
Thanx.
Nvd.

2. Originally Posted by beezee99
Hi all. It may be simple, but I do not seem to recall it. How can I solve the following

As an example u may like to set n to any integer, say, 5.
Thanx.
Nvd.
(it doesn't make sense to have x in the limits when it is the variable you're integrating with respect to...EDIT, if it is pi, then no worries )

the way to handle an integral like this, is to simply expand the numerator, using the binomial theorem if n is large enough to make it useful, and then dividing each term by x. for the first term, the 1/x, the integral would be ln|x|, for the others, use the power rule for integrals: $\displaystyle \int x^n ~dx = \frac {x^{n + 1}}{n + 1} + C$

then apply the fundamental theorem of calculus to deal with the limits on the integral.

3. Originally Posted by beezee99
Hi all. It may be simple, but I do not seem to recall it. How can I solve the following

As an example u may like to set n to any integer, say, 5.
Thanx.
Nvd.

It's hard to read what the upper limit is, but let's supose it is pi, so:

$\displaystyle{\int\limits^\pi_0\frac{(1-x)^n}{x}\,dx=\int\limits^\pi_0\frac{\sum\limits^n_ {k=0}\binom{n}{k}(-x)^k}{x}\,dx=\int\limits^\pi_0\left(\frac{1}{x}-n+\frac{n(n-1)}{2}x-\ldots\pm x^{n-1}\right)\,dx$ ,

and now use linearity of the integral to solve each of the above...

Tonio

4. $\int \frac{(1-x)^n}{x} ~dx$ where $n$ is natural number .

Sub. $x = 1 -t$

$dx = -dt$

We have $- \int \frac{t^n}{1-t}~dt$

$= - \int \frac{ 1 - (1-t^n) }{1-t}~dt$

$= - \int \left[ \frac{1}{1-t} - (1+ t + t^2 + ... + t^{n-1} ) \right] ~dt$

$= \frac{t}{1} + \frac{t^2}{2} + \frac{t^3}{3} + .... + \frac{t^n}{n} + \ln|1-t| + C$

or $= \frac{(1-x)}{1} + \frac{(1-x)^2}{2} + \frac{(1-x)^3}{3} + ... + \frac{(1-x)^n}{n} + \ln|x| + C$

5. Hi guys. Sorry, the upper limit is 1.

6. It's the lower limit, 0, that is the problem!

7. So then how to resolve the issue for lower limit of 0 ?

8. Originally Posted by beezee99
So then how to resolve the issue for lower limit of 0 ?
it's an improper integral then. you would have to compute:

$\displaystyle \lim_{N \to 0^+} \int_N^1 \frac {(1 - x)^n}x~dx$

(the integral doesn't converge...)

9. How about if I have the following where m and n are different integers?

Will the ln(x) terms cancel out?
Thanx.

10. Originally Posted by beezee99
How about if I have the following where m and n are different integers?

Will the ln(x) terms cancel out?
Thanx.
No, both integrals would be considered separately. you apply the fundamental theorem one at a time since the limits are different

11. Hi. The limits are the same i.e. 0<=x<=1 for both the integrals. The only difference between the two is the powers of (1-x). So?

12. Originally Posted by beezee99
Hi. The limits are the same i.e. 0<=x<=1 for both the integrals. The only difference between the two is the powers of (1-x). So?
well in that case, you can put both under the jurisdiction of a single integral, and then yes, the 1/x 's will cancel. so you will have a convergent integral.