Dealing with functions of several variables

Nick89
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Homework Statement


Assuming x > -1 and y > -1, find the following integral:
\displaystyle \int_0^1 \frac{t^x - t^y}{\ln t} dt


Homework Equations





The Attempt at a Solution


I have no idea where to start on this one... It came up in an exam for a class dealing with functions of several variables (eg, f(x,y,z)) so I don't think it's a usual substitution problem...

I tried writing it as \int t^x \ln(-t) dt (minus the same for y) but I couldn't get any further...

Do I perhaps have to find some taylor polynomial or something?
 
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Nick89 said:
I tried writing it as \int t^x \ln(-t) dt (minus the same for y) but I couldn't get any further...
This is wrong that's not the same integral ( do you know why?)

Use IBP.
 


Sorry I screwed up there, heh.

Anyway, I just found out that this question in the exam was part of some chapter that is now omitted, that would explain why I have no idea how to start on this one ^^
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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