Challenging improper & definite integral

In summary, the conversation is about evaluating two integrals and the speaker is asking if there is a more elegant way to solve them and if it is possible to evaluate them other than numerically. They also mention that they already know the answers and want to see what others would say. The conversation ends with the speaker providing an elegant solution for the second integral.
  • #1
mathwizarddud
25
0
Evaluate

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

and

[tex]\int_0^1\frac{x-1}{\ln\, x} \; dx [/tex]
 
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  • #2
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?
 
  • #3
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
:rofl:
 
  • #4
mathwizarddud said:
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
:rofl:
Yeah okay no one cares.
 
  • #5
mathwizarddud said:
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
:rofl:

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?
 
  • #6
Elegant solution for the second integral (at least in my opinion):

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

Consider [tex]\int^b_a dt \int^1_0 x^t dx[/tex].

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

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

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

--------
EDIT: Whoops I forgot to evaluate the actual integral here. Letting b=1, a=0 gives us the original integral as equal to [itex]\log_e 2[/itex].
 
  • #7
Gib Z said:
Elegant solution for the second integral (at least in my opinion):

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

Consider [tex]\int^b_a dt \int^1_0 x^t dx[/tex].

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

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

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

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

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

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

1. What is the definition of a definite integral?

A definite integral is a mathematical concept used to calculate the area under a curve on a graph. It involves finding the limit of a sum of infinitely small rectangles that approximate the shape of the curve, and can be thought of as the signed area between the curve and the x-axis.

2. Can any function be integrated using definite integrals?

Yes, any continuous function can be integrated using definite integrals. However, some functions may require more advanced techniques such as substitution or integration by parts.

3. What is the difference between an improper integral and a definite integral?

A definite integral has both upper and lower limits, indicating the specific range over which the integration is being performed. An improper integral has either one or both of its limits at infinity, or the function being integrated is unbounded or discontinuous at one or more points within the limits.

4. How do you solve an improper integral?

To solve an improper integral, you must first determine if it is convergent or divergent. If it is convergent, you can use the limit comparison test, direct comparison test, or the integral test to evaluate the integral. If it is divergent, you can use the limit comparison test or the direct comparison test to show that the integral does not converge.

5. Why are improper integrals important in mathematics?

Improper integrals are important in mathematics because they allow us to extend the concept of integration to functions that may not be defined or continuous over a finite interval. They also have many applications in physics, engineering, and other areas of science.

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