Challenging improper & definite integral

[mathwizarddud]

Evaluate

\int_{0}^{\infty} x\ (e^{3x}-1)^{-1/3}\ dx

and

\int_0^1\frac{x-1}{\ln\, x} \; dx

 

[HallsofIvy]

Why? I presume you have some very good reason for wanting to do those integrals, perhaps because you need to learn how to do that sort of thing. In that case, it is far better for you to show what you have done so far so we can make suggestions on how to continue.

And if this is not homework, do you have any reason to believe it is possible to evaluate those other than numerically?

 

[mathwizarddud]

This is not homework of course. I know the answers already (I made one of them up), and I just want to see what would people say if they don't exactly know how to solve it (by hand) + if there's a more elegant way to solve these two so that I could learn something, too! LOL

 

[rocomath]

This is not homework of course. I know the answers already (I made one of them up), and I just want to see what would people say if they don't exactly know how to solve it (by hand) + if there's a more elegant way to solve these two so that I could learn something, too! LOL

Yeah okay no one cares.

 

[dirk_mec1]

This is not homework of course. I know the answers already (I made one of them up), and I just want to see what would people say if they don't exactly know how to solve it (by hand) + if there's a more elegant way to solve these two so that I could learn something, too! LOL

Hey I got a better idea why don't you show us how you solved them and then maybe we'll tell you if there's a shorter way home, m'kay?

 

[Gib Z]

Elegant solution for the second integral (at least in my opinion):

Identity: \int^b_a dx \int^t_{t_0} f(x,t) dt = \int^t_{t_0} dt \int^b_a f(x,t) dx

Consider \int^b_a dt \int^1_0 x^t dx.

Evaluating directly, we get \int^b_a \frac{1}{t+1} dt = \log_e \left( \frac{b+1}{a+1} \right).

Applying the identity, we get \int^1_0 dx \int^b_a x^t dt, and evaluating the inner integral shows that is equal to \int^1_0 \frac{x^b -x^a}{\log_e x} dx.

Hence, \int^1_0 \frac{x^b-x^a}{\log_e x} dx = \log_e \left( \frac{b+1}{a+1} \right)

--------
EDIT: Whoops I forgot to evaluate the actual integral here. Letting b=1, a=0 gives us the original integral as equal to \log_e 2.

 

[mathwizarddud]

Elegant solution for the second integral (at least in my opinion):

Identity: \int^b_a dx \int^t_{t_0} f(x,t) dt = \int^t_{t_0} dt \int^b_a f(x,t) dx

Consider \int^b_a dt \int^1_0 x^t dx.

Evaluating directly, we get \int^b_a \frac{1}{t+1} dt = \log_e \left( \frac{b+1}{a+1} \right).

Applying the identity, we get \int^1_0 dx \int^b_a x^t dt, and evaluating the inner integral shows that is equal to \int^1_0 \frac{x^b -x^a}{\log_e x} dx.

Hence, \int^1_0 \frac{x^b-x^a}{\log_e x} dx = \log_e \left( \frac{b+1}{a+1} \right)

--------
EDIT: Whoops I forgot to evaluate the actual integral here. Letting b=1, a=0 gives us the original integral as equal to \log_e 2.

Yes, double integration is exactly what I had in mind.
Try the first one; it's a bit harder.

The answer is: (ln(3)+ \pi/3^{1.5})*\pi/3^{1.5}.