Difficult improper integral using mathematica

[Drezzan]

Homework Statement

evaluate $$\int_0^1\frac{Ln(x)}{1+x}\,dx$$

Homework Equations

I know the way to solve most improper integrals; replacing 0 or the bound causing the issue with a variable and have the limit of the integral as the variable goes to infinity. My question is using mathematica how do I get a reasonable result?

The Attempt at a Solution

I had started just looking for the indefinite integral first, just to keep an eye on the method mathematica was using and making sure I agreed. I had not tried this by hand yet. $$\int\frac{Ln(x)}{1+x}\,dx$$
This however spat out an crazy result about a screen length wrong with an IF statement and an assumptions (it is far to long to type here). I need help at this juncture. What caused the issue and how do I resolve it to allow me to take the limit of the integral?

 

[phyzguy]

By Ln(x) do you mean the natural log of x ( usually written ln(x) )? If so, Mathematica gives me a simple result:

Integrate[Log[x]/(1 + x), {x, 0, 1}] = - (pi^2 / 12)

 

[Ray Vickson]

Homework Statement

evaluate $$\int_0^1\frac{Ln(x)}{1+x}\,dx$$

Homework Equations

I know the way to solve most improper integrals; replacing 0 or the bound causing the issue with a variable and have the limit of the integral as the variable goes to infinity. My question is using mathematica how do I get a reasonable result?

The Attempt at a Solution

I had started just looking for the indefinite integral first, just to keep an eye on the method mathematica was using and making sure I agreed. I had not tried this by hand yet. $$\int\frac{Ln(x)}{1+x}\,dx$$
This however spat out an crazy result about a screen length wrong with an IF statement and an assumptions (it is far to long to type here). I need help at this juncture. What caused the issue and how do I resolve it to allow me to take the limit of the integral?

Maple gets a nice, simple formula (but involving the non-elementary "dilog" function):
f:=log(x)/(1+x);

ln(x)/(1+x)

J:=int(f,x);

J := dilog(1 + x) + ln(x) ln(1 + x)

We can evaluate the definite integral from this:

J1:=limit(J,x=1,left);

-Pi^2 /12

J0:=limit(J,x=0,right);

J0 := 0

Therefore, the definite integral is - π²/12. We also get this right away if we ask Maple to do the definite integral.

I do not have access to Mathematica, so cannot help you with that aspect of your problem.

RGV

 

[Drezzan]

I am still unsure of what I did wrong but after retyping it today I got the same answer. So I don't have to do any other operations even though it is in improper form? I just want to make sure that there wouldn't be a different result if preformed a different way.

 

[Ray Vickson]

I am still unsure of what I did wrong but after retyping it today I got the same answer. So I don't have to do any other operations even though it is in improper form? I just want to make sure that there wouldn't be a different result if preformed a different way.

As I said already, I get -pi^2/12 when I do it in Maple. I do not have access to Mathematica, but when I try it on Wolfram Alpha I get a numerical answer which is, in fact, the numerical value of -pi^2/12. I am assuming that Mathematica is the computational engine behind Wolfram Alpha, so Mathematica knows how to do the integral and gets the right answer. The only conclusion I can draw is that something is wrong with your input. But, not being a Mathematica user, I cannot tell what your problem may be. Why don't you post the detailed instruction here, so others can look at it?

RGV