Calculating integral by differentiating first

physmatics
Messages
16
Reaction score
0

Homework Statement


Calculate the integral
I = \int (t^x - 1)/ln(t) dt, boundaries: 0 \leq t \leq 1, x \geq 0
by differentiating first with respect to x.

Homework Equations


-

The Attempt at a Solution


I have no idea how to solve this, but it's on our sample exam and there are no solutions... =/ Differentiating with respect to x gives me:
d/dx((t^x - 1)/ln(t)) = ln(t)e^(xln(t))/ln(t) + 0 = e^(xln(t)) = t^x
Can I use this in any way? Maybe substitute t^x in the integral with d/dx(t^x - 1)/ln(t)?
Any clues appreciated!

 
Physics news on Phys.org
Remember that the Leibniz Rule gives you

\frac{d}{dx}\int_0^1{\frac{t^x-1}{ln(t)}dt}=\int_0^1{\frac{d}{dx}\left(\frac{t^x-1}{ln(t)}\right)dt}=\int_0^1{t^xdt}

Now, the right-hand side looks like an easy integral...
 
Very nice. Physmatics, becareful to distinguish between the derivatives of x^a and a^x.
 
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...
Back
Top