How can I evaluate the integral of x/lnx if all methods seem to fail?

[Xyius]

For some reason, I can not seem to evaluate this integral.

\int \frac{x}{ln(x)} dx

When I plug it into maple it says the solution is

\frac{1}{2} \frac{x^2}{ln(x)}

I cannot seem to get it for some reason! I have tried every integration technique in the book. :\

 

[Mute]

That is not a correct result, as you can check by differentiating. It looks like you accidentally wrote log(y) or something which mathematica treated as a constant, and so it gave you the integral of x. The real result involves the exponential integral,

\mbox{Ei}(2\ln(x))

http://www.wolframalpha.com/input/?i=integrate+x/ln(x)

 

[Xyius]

That is not a correct result, as you can check by differentiating. It looks like you accidentally wrote log(y) or something which mathematica treated as a constant, and so it gave you the integral of x. The real result involves the exponential integral,

\mbox{Ei}(2\ln(x))

http://www.wolframalpha.com/input/?i=integrate+x/ln(x)

You were right, I forgot to put the parenthesis on the natural logarithm. I haven't learned what an exponential integral is yet. I am currently in Differential Equations but "Ei" doesn't look familiar. :\

 

[Mute]

You were right, I forgot to put the parenthesis on the natural logarithm. I haven't learned what an exponential integral is yet. I am currently in Differential Equations but "Ei" doesn't look familiar. :\

The Exponential Integral funtion, \mbox{Ei}(x), is a special function defined by

\mbox{Ei}(x) = \int_{-\infty}^x dt~\frac{e^t}{t}

To be fair, this is just obtained by making a change of variables on your original integral; however, this is a special function that should be available in mathematica or matlab, so it is a standard-ish form you can use.

 

[Xyius]

The Exponential Integral funtion, \mbox{Ei}(x), is a special function defined by

\mbox{Ei}(x) = \int_{-\infty}^x dt~\frac{e^t}{t}

To be fair, this is just obtained by making a change of variables on your original integral; however, this is a special function that should be available in mathematica or matlab, so it is a standard-ish form you can use.

I did get that form quite a few times when I was trying to solve it. But I stopped at that point because it seemed like a dead end. So would that mean that the solution to the integral is an integral? Would this suggest that there is no actual solution to it? I guess you would use approximation techniques at that point to evaluate it. Am I correct with this assumption?

 

[Anonymous217]

I haven't looked at it closely, but it just looks like the integration by parts using u = x and setting dv as the logarithmic integral.

 

[Mute]

I did get that form quite a few times when I was trying to solve it. But I stopped at that point because it seemed like a dead end. So would that mean that the solution to the integral is an integral? Would this suggest that there is no actual solution to it? I guess you would use approximation techniques at that point to evaluate it. Am I correct with this assumption?

Your integral is one that can't be expressed in elementary functions. However, while the exponential integral function is still defined as an integral, it is a standard function that is recognized by many computer math systems, and so this form is more useful than the original integral form. The programs will use a variety of techniques to evaluate the function for different regimes of x, but I do not the specific details.