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

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In summary: It is possible that the exponential integral function itself is not "elementary".In summary, The problem discussed is evaluating the integral \int \frac{x}{ln(x)} dx, which has a solution of \frac{1}{2} \frac{x^2}{ln(x)} according to Maple. However, this is not the correct result as it does not pass the differentiation test. The actual solution involves the exponential integral function, \mbox{Ei}(2\ln(x)), which is defined as \mbox{Ei}(x) = \int_{-\infty}^x dt~\frac{e^t}{t}. While this form still involves an integral, it is a standard function recognized by many
  • #1
Xyius
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For some reason, I can not seem to evaluate this integral.

[tex]\int \frac{x}{ln(x)} dx[/tex]

When I plug it into maple it says the solution is

[tex]\frac{1}{2} \frac{x^2}{ln(x)}[/tex]

I cannot seem to get it for some reason! I have tried every integration technique in the book. :\
 
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  • #2
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,

[tex]\mbox{Ei}(2\ln(x))[/tex]

http://www.wolframalpha.com/input/?i=integrate+x/ln(x)
 
  • #3
Mute said:
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,

[tex]\mbox{Ei}(2\ln(x))[/tex]

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. :\
 
  • #4
Xyius said:
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, [itex]\mbox{Ei}(x)[/itex], is a special function defined by

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

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.
 
  • #5
Mute said:
The Exponential Integral funtion, [itex]\mbox{Ei}(x)[/itex], is a special function defined by

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

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?
 
  • #6
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.
 
  • #7
Xyius said:
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.
 

What is the integral of x/lnx?

The integral of x/lnx is equal to x(lnlnx - 1) + C, where C is the constant of integration.

How do you solve the integral of x/lnx?

To solve the integral of x/lnx, you can use the substitution method by letting u = lnx. This will result in the integral becoming ∫u du, which is easy to solve. After solving for u, you can substitute back in lnx for u to get the final answer.

What is the domain of the integral of x/lnx?

The domain of the integral of x/lnx is (0, ∞). This means that the function is only defined and can only be integrated for positive values of x.

Is the integral of x/lnx a continuous function?

Yes, the integral of x/lnx is a continuous function. This means that the function is defined and has a value for every point on its domain, and there are no sudden jumps or breaks in the graph of the function.

Can the integral of x/lnx be used to solve real-world problems?

Yes, the integral of x/lnx can be used to solve real-world problems in various fields such as physics, economics, and engineering. It can be used to calculate the area under a curve, which can represent physical quantities such as displacement, velocity, or acceleration. It can also be used to find the average value of a function, which is useful in determining averages in various fields.

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