Incredibly Hard Integration..or not... (1 Viewer)

[Slidey]

hhmmmm..good point, but still, if i siad solve for x: lnx=0, there is no solution why???because ln0 is undefined..anyway this is irrelevant does anybody out there have a solution!?!?!?!

Actually, the solution is x=1.

lnx=0
e^(lnx)=e^0
x=1

Hmm. While we're on the topic of integration, exponentials and logarithms...

∫ dx/(e^x-e^-x)

I guess you'd call it one of those ones where you either get it or you don't. Try it, though - it's not hard.

 

[m_isk]

oooopppss!! i meant ln0 is undefined

 

[lucifel]

∫dx/(e^x-e^-x)

= ∫dx/ (e^x - (1/e^x))

= ∫dx / (e^2x - 1)/e^x

= ∫ e^x dx / (e^2x - 1)

put u = e^x, du = e^x dx

thus Integration is:

∫du / (u^2 -1 )

= 1/2 ln [(u-1)/(u+1)] + C (this step can be either treated as a standard integral, or it is done by partial fractions, either or really.)

= 1/2 ln [(e^x - 1)/(e^x +1)] + C

 

[jumb]

ln²x²/2 + c?

The integrator agrees with you.

http://integrals.wolfram.com/

 

[lucifel]

The integrator agrees with you.

http://integrals.wolfram.com/

but wolfram doesn't accept 'ln', it only accepts log, so therefore that integral is wrong.

The one that ngai posted is prolly correct, I don't think this is a 4U integration anymore.

 

[Slidey]

Re the e^x integral: correct!

The one that ngai posted is correct (and is also the one the integrator gives - rightly so since the integrator runs on Mathematica)! But it also includes an integral of its own, so it's not that helpful (if you wanted to use Simpson's rule or something, you could just as easily apply it to the original integral).

 

[dawso]

maple says its ln(ln(x))*x+Ei(1,-ln(x))
whatever Ei is....

i started reading this thread and the got to this post, cant be bothered readin any more, if ngai dont know it, dw m8...

 

[colless]

maple says its ln(ln(x))*x+Ei(1,-ln(x))
whatever Ei is....

I realise this thread is over now, but just letting those who read it know that Ei is knows as the exponential integral. The function is rather messy, well actually very messy and it simply cannot be broken down under further analytical methods into smaller functions, i.e. logs or polynomials. This question annoyed me for quite some time, but after using mathematica, and talking to a mathematician I am friends with I completely understand....that i am out of my depth lol. I could give a brief summary of how you arrive at Maple's answer but i think it would be mostly unintelligble to most people as it is way outside the scope of the course and uses some rather interesting methods. In summary, interesting problem to note, but dont worry about getting bogged down trying to work it out.

 

[Xenocide]

Bleh, posted from wrong username, but I think you get the gist, basically Ngai's answer is correct, though to get at the answer you need some rather advanced methods.

 

[Slidey]

Is this exponential integral solvable by non-approximate methods? If not, why not just use integration by parts to arrive at, as mentioned previously:

xln(ln(x)) - integral 1/ln(x)

And for the integral part apply an approximate method?