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Integration by iteration

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?

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Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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]1e= e ln e - 1 ln 1 = e.
1e (ln x)k dx = e - k ∫1e (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.


Thanks a ton Andre. You confirmed that I was doing it correctly. I realize now that I lost sight of what I ultimately had to do and thought I needed to be able to evaluate the resulting integral. When I saw that it wasn't simplified, I thought I was doing the mechanics wrong. You were right on. I used the equality to make a reduction formula and set up a small table of values. I appreciate the course correction.
Glad I could help, Judith. Iterated integrals are fun and very important. You can use them to define the factorial (k!) for any real number, not just positive integers! You can even use them to define a fractional derivative, e.g., the 1/2-th derivative of ln(x). Strange and beautiful.
William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
I hope I can help you with this, Judith.
If k = 1
x =1x=e ln(x)dx = x[ln(x)] - x = 1
If k = 2
x =1x=e [ln(x)]2dx =2x + x[ln(x)]- 2xln(x) = e - 2
If k = 3
x =1x=e [ln(x)]3dx = 6 - 2e
x =1x=e [ln(x)]4dx = 9e - 24
x =1x=e [ln(x)]5dx = 120 - 44e
I'm not sure if this is exactly what you need.  If not, please write back.


Thanks William for your speedy response. It wasn't the method I was required to use, but I have it covered now.