Indefinite integral of 1/x times natural log of (a + bx).
a and b are real numbers.

Dear Jakearle,

Thanks for your question.  An excellent resource for computing
indefinite integrals is the following Integrator from Wolfram
Reseaech:

http://integrals.wolfram.com/

The site fills in a random sample, which you can change to anything
you like.  For infoormation on how to enter a formula, see these tips:

http://integrals.wolfram.com/about/input/

The input format is, not surprisingly, the same as that for
Mathematica, if you are familiar with this software.

The input is the following:
(1/x) / Log[a+bx]

Using this tool, the integral is
(Log x) / Log (a+bx)

(Log here is the natural log, aka ln.)


Another integral lookup tool can be found here, at Berkeley:
http://torte.cs.berkeley.edu:8010/tilu


I hope this was helpful.

Best,

      -welte-ga

---

Clarification of Answer by welte-ga on 14 May 2005 07:27 PDT

My apologies... I misunderstood the statement of your question.  The
input should be:

(Log[a+bx])/x

The resulting integral is then:

Log[a+bx] * Log[x]


         -welte-ga

---

Clarification of Answer by welte-ga on 14 May 2005 07:54 PDT

OK... Must not be my day.  Don't try to do math before coffee!


After playing around with the Wolfram integration tool, I realized
that it doesn't recognize implicite multiplication without a space
between the variables.  For this reason, in input I gave you, "bx" was
treated as a constant, rather than as b times x.  This can be fixed by
either putting a space in (b x) or making the multiplication explicite
(b*x).  The correct input should be:

(Log[a+b*x])/x

This results in a more complicated indefinite integral, as one might expect:

Log[x] (Log[a+b*x] - Log[1+b*x/a]) - PolyLog[2, -bx/a].

The PolyLog (Polylogarithm) function, also known as Jonquière's
function, is described in detail here:
http://mathworld.wolfram.com/Polylogarithm.html

and here:
http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/


Using traditional notation, then, one can substitute

PolyLog[2, -b x/a] = Li_2 [-b x/a]

Also, Li_2 [-b x/a] can be written as

Sum (from k=1 to infinity) [((-b x/a)^k)/k^2]


Sorry for the confusion.  Let me know if you require any clarification.

        -welte-ga

---

Request for Answer Clarification by jakearle-ga on 15 May 2005 16:03 PDT

Shouldn't that simplify to ln(a)*ln(x) - PolyLog[2, -bx/a], 
since ln(1 + bx/a) = ln[(a + bx)/a] = ln(a + bx) - ln(a)?

---

Clarification of Answer by welte-ga on 15 May 2005 19:56 PDT

That is correct.

        -welte-ga