<p>How do you integrate 1/ln(x)... I'm having brain cramps</p>
<p>li(x) :)</p>
<p>You have stumbled upon the logarithmic integral. Integrate it using elementary means, and you'll be famous.</p>
<p>umm.......will by parts method work>> sorry, have not touched maths since finished a levels last nov so a bit rusty yeah....
by parts is the only thing i remember or maybe some kinda substitution.</p>
<p>It is certainly something out of my current knowledge of integration :) I've tried to solve the same sort of things, i.e int(e^(x^2), x) but it has proven to be impossible at this state of education.</p>
<p>I don't know if it can be expressed as a finite sum of elementary functions. I will have to check.</p>
<p>1/ln(x) is somewhat like sin(x)/x, which can be integrated, but only into a series, I believe. I know that if you convert sin(x) into a series, then divide each term by x, you can integrate each term separately. But I dunno if there's an analogous method of integrating 1/ln(x), and even if there were, the series would be in the denominator so the fraction couldn't be split. </p>
<p>Perhaps you could introduce an intermediary variable b, to represent 1/ln(x) in terms of e and b, and then you could substitute with the definition of e, and then integrate...? Just a thought.</p>
<p>Check out mathworld: <a href="http://mathworld.wolfram.com/NaturalLogarithm.html%5B/url%5D">http://mathworld.wolfram.com/NaturalLogarithm.html</a></p>
<p>Good luck.</p>
<p>There is a trick that works with integral e^(x^2) dx from 0 to infinity where you square the integral and convert to polar coordinates. The value you get is rt(pi)/2 I think.</p>
<p>I think you can do it by parts.</p>
<p>amrik, you are correct on that one. int(e^(x^2),x, -inf, inf) = sqrt(pi)</p>
<p>Uh... if you just wanted the basic formula,
integral (ln ax)^n dx = x [(ln ax)^n] - n [integral (ln ax)^(n-1) dx]
n = -1</p>
<p>Which, of course, doesn't do anything for non-18 people, but :shrug: It's the closest I can get without going into the stuff the other websites have discussed.</p>
<p>About the int(e^(x^2), x=0..infinity);</p>
<p>You are right :) Just got the method from Google search and it is quite surprising as it subtly uses the bi-intergration, which I just slightly touched a year ago (God, I am relieved of my inabilities now :D)</p>
<p>oh man, is this called multivariable calculus?? and i thought Alevel were headache :p</p>
<p>I'm glad I got over having fun with these functions that when integrated, can't be expressed as finite sums of elementary functions. If you're curious about these types of functions, Mathematica and Maple should support them. I did once find a nested series for the integral of x^x dx, but it only converged from 0 to 1.</p>
<p>My question for Kirbus is: did you need this for a class or for a physical application or something... or just curious?</p>
<p>I did it in maple and got -Ei(1,-ln(x)), brief inspection indicated that this was the same as Li(x) + C | C = i*pi, where i^2 = -1.</p>
<p>Sherwin: I actually miscopied a problem. It involved integrating 1/(xln(x)). I should really doublecheck that stuff... In the end I was just curious though.</p>
<p>... 1/(xln(x)) is a completely different story... </p>
<p>lol</p>
<p>heh</p>
<p>that's just ln(ln(x))</p>
<p>Yeah, i kicked myself pretty hard after that. I'm ok now, although my brain hurts and I have pages of work that gets me nowhere...</p>