Integration by iteration

Question

Calculate an expression for the int from 1 to e of (ln(x))^k dx in terms of the integral from 1 to e of (ln(x))^k-1 dx.

essentially, I need to integrate by parts to get this equality:

The integral from 1 to e of (ln(x))^k dx= (e-k) times(integral from 1 to e of (ln(x))^k-1  dx.

Then I will take the value of the original integral with k=1 to help me form a table of values for a series of k's. 

I get close, but can't get there. Anyone able to help?

Answer 1

Remember the integration-by-parts formula:

∫u v' dx = u v - ∫ v u' dx.

 

Now let u=(ln x)^k and v'=1. Then u'=k (ln x)^k-1/x and v=x. Put this into the formula and you get:

 

∫(ln x)^k dx = x (ln x)^k - k ∫(ln x)^k-1 dx.

 

For the definite integral from 1 to e, evaluate the integrated term at those boundaries:

[x (ln x)^k]₁^e= e ln e - 1 ln 1 = e.

 

Therefore,

∫₁^e (ln x)^k dx = e - k ∫₁^e (ln x)^k-1 dx.

 

Note: the iteration holds for any real number k≠0; however, you will get a finite number of terms if and only if k is a positive integer.

Answer 2

I hope I can help you with this, Judith.

 

If k = 1

 

∫_{x =1}^x=e ln(x)dx = x[ln(x)] - x = 1

 

If k = 2

 

∫_{x =1}^x=e [ln(x)]²dx =2x + x[ln(x)]² - 2xln(x) = e - 2

 

If k = 3

 

∫_{x =1}^x=e [ln(x)]³dx = 6 - 2e

 

∫_{x =1}^x=e [ln(x)]⁴dx = 9e - 24

 

∫_{x =1}^x=e [ln(x)]⁵dx = 120 - 44e

 

I'm not sure if this is exactly what you need.  If not, please write back.