1. ## Challenging improper integral

Prove that

$\int_0^{\infty}x(exp(3x)-1)^{-1/3} dx
=(ln(3)+ \pi/3^{1.5})*\pi/3^{1.5}
$

2. Originally Posted by mathwizard
Prove that

∫dx*x*(exp(3x) –1)^(-1/3) from x=0 to infinity =(ln(3)+ pi/3^1.5)*pi/3^1.5
Considering the integral I would assume the rate of convergence would be very high, and numerically

$\int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx\ne\bigg(\ln(3)+\frac{\pi}{3^{\ frac{3}{2}}}\bigg)\frac{\pi}{3^{\frac{3}{2}}}$

are you sure this is correct? I am sure you are but I am just wondering

3. I just glanced at it. Gotta go for a little while. But, maybe, just maybe, we can transform it into a beta function form and go from there.

$B(p,q)=\int_{0}^{\infty}\frac{y^{p-1}}{(1+y)^{p+q}}dy$

With the proper subs, this may be doable. Give it a whirl.

4. Originally Posted by mathwizard
Prove that

$I=\int_0^{\infty}\frac{x}{\sqrt[3]{e^{3x}-1}} dx
=\frac{\pi}{3\sqrt{3}}\left(\ln 3 + \frac{\pi}{3\sqrt{3}}\right)
$
first two simple facts that we will use them later on in the solution:

$(*)$ if $y > 0$, then $\int_0^{\infty}\frac{dt}{t^3+y}=\frac{1}{\sqrt[3]{y^2}} \int_0^{\infty}\frac{dt}{1+t^3},$ and $\int_0^{\infty} \frac{dt}{1+t^3} = \frac{2\pi}{3\sqrt{3}}.$

now, in your integral, put $e^{3x}-1=\frac{1}{t^3}.$ then $x=\frac{1}{3}\ln\left(1+\frac{1}{t^3}\right),$ and $dx=\frac{-dt}{t(1+t^3)}.$ thus:

$I=\frac{1}{3}\int_0^{\infty}\frac{\ln(1+\frac{1}{t ^3})}{1+t^3} \ dt=\frac{1}{3}\int_0^{\infty} \int_0^1\frac{1}{(t^3+y)(t^3+1)} \ dy \ dt$

$=\frac{1}{3} \int_0^1 \int_0^{\infty} \frac{1}{(t^3+y)(t^3+1)} \ dt \ dy=\frac{1}{3}\int_0^1 \frac{1}{1-y}\int_0^{\infty} \left(\frac{1}{t^3+y}-\frac{1}{t^3+1}\right) \ dt \ dy.$ thus by $(*)$:

$I=\frac{2\pi}{9\sqrt{3}}\int_0^1 \frac{1-\sqrt[3]{y^2}}{\sqrt[3]{y^2}(1-y)} \ dy.$ now let $y=z^3.$ then we will have:

$I=\frac{2\pi}{3\sqrt{3}} \int_0^1 \frac{1+z}{1+z+z^2} \ dz=\frac{2\pi}{3\sqrt{3}}\left[\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z+1}{\sqrt{3}}\right)+\frac{1}{2}\l n(1+z+z^2) \right]_ 0^1$

$=\frac{\pi}{3\sqrt{3}}\left(\ln 3 + \frac{\pi}{3\sqrt{3}} \right). \ \ \ \ \square$

5. NCA, you're just a show off

Beautiful solution. Do you think we could manipulate it somehow and use the Beta as I mentioned?. I gave it a short go, but failed to arrive at anything tangible. You probably can though.

6. Originally Posted by NonCommAlg
first two simple facts that we will use them later on in the solution:

$(*)$ if $y > 0$, then $\int_0^{\infty}\frac{dt}{t^3+y}=\frac{1}{\sqrt[3]{y^2}} \int_0^{\infty}\frac{dt}{1+t^3},$ and $\int_0^{\infty} \frac{dt}{1+t^3} = \frac{2\pi}{3\sqrt{3}}.$

now, in your integral, put $e^{3x}-1=\frac{1}{t^3}.$ then $x=\frac{1}{3}\ln\left(1+\frac{1}{t^3}\right),$ and $dx=\frac{-dt}{t(1+t^3)}.$ thus:

$I=\frac{1}{3}\int_0^{\infty}\frac{\ln(1+\frac{1}{t ^3})}{1+t^3} \ dt=\frac{1}{3}\int_0^{\infty} \int_0^1\frac{1}{(t^3+y)(t^3+1)} \ dy \ dt$

$=\frac{1}{3} \int_0^1 \int_0^{\infty} \frac{1}{(t^3+y)(t^3+1)} \ dt \ dy=\frac{1}{3}\int_0^1 \frac{1}{1-y}\int_0^{\infty} \left(\frac{1}{t^3+y}-\frac{1}{t^3+1}\right) \ dt \ dy.$ thus by $(*)$:

$I=\frac{2\pi}{9\sqrt{3}}\int_0^1 \frac{1-\sqrt[3]{y^2}}{\sqrt[3]{y^2}(1-y)} \ dy.$ now let $y=z^3.$ then we will have:

$I=\frac{2\pi}{3\sqrt{3}} \int_0^1 \frac{1+z}{1+z+z^2} \ dz=\frac{2\pi}{3\sqrt{3}}\left[\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z+1}{\sqrt{3}}\right)+\frac{1}{2}\l n(1+z+z^2) \right]_ 0^1$

$=\frac{\pi}{3\sqrt{3}}\left(\ln 3 + \frac{\pi}{3\sqrt{3}} \right). \ \ \ \ \square$
How hard is it to do difficult definite integrals prior to learning double and triple integration, because it seems very hard

7. Originally Posted by Mathstud28
Considering the integral I would assume the rate of convergence would be very high, and numerically

$\int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx\ne\bigg(\ln(3)+\frac{\pi}{3^{\ frac{3}{2}}}\bigg)\frac{\pi}{3^{\frac{3}{2}}}$

are you sure this is correct? I am sure you are but I am just wondering
The difference between the two is only $4.5 \times 10^{-10}~\%$! Why would you say this is so bad an estimate??

-Dan

8. Originally Posted by topsquark
The difference between the two is only $4.5 \times 10^{-10}~\%$! Why would you say this is so bad an estimate??

-Dan
My 89 gives .18277 and 1.02976?

9. Originally Posted by Mathstud28

How hard is it to do difficult definite integrals prior to learning double and triple integration, because it seems very hard
It's not that hard, it's about practice, lot of practice and learn everyday parameters to construct double integrals.

I like this way to tackle problems like this one, besides being a nice way to do it, you simplify a lot nasty calculations.

10. Originally Posted by Krizalid
It's not that hard, it's about practice, lot of practice and learn everyday parameters to construct double integrals.

I like this way to tackle problems like this one, besides being a nice way to do it, you simplify a lot nasty calculations.
By everday paramaters do you mean

$\arctan(x)=\int_0^x\frac{dy}{1+y^2}$

or

$\ln(x)=\int_1^x\frac{dy}{y}$

?

11. Originally Posted by Mathstud28

By everday paramaters do you mean

$\arctan(x)=\int_0^x\frac{dy}{1+y^2}$

or

$\ln(x)=\int_1^x\frac{dy}{y}$

?
Yeah, but those are basic ones, you have to find/learn lots of them. (Of course, it depends of your integration skills how to use them to tackle a problem.)

12. Originally Posted by Mathstud28
My 89 gives .18277 and 1.02976?
$\int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx = 1.0297616606482$

$\bigg(\ln(3)+\frac{\pi}{3^{\frac{3}{2}}}\bigg)\fra c{\pi}{3^{\frac{3}{2}}} = 1.0297616606528$

For a difference of $4.6 \times 10^{-12}$ (or so.)

-Dan

13. Originally Posted by topsquark
$\int_0^{5000}\frac{x}{(e^{3x}-1)^{\frac{1}{3}}}dx = 1.0297616606482$

$\bigg(\ln(3)+\frac{\pi}{3^{\frac{3}{2}}}\bigg)\fra c{\pi}{3^{\frac{3}{2}}} = 1.0297616606528$

For a difference of $4.6 \times 10^{-12}$ (or so.)

-Dan
::grumbles:: stupid 89

14. Originally Posted by Mathstud28
::grumbles:: stupid 89
Are you sure you put it in right? My 89 gave me the correct (approximate) answer. Perhaps you forgot the parentheses around the $\frac13$ or the $3x$?

Edit: Ah, it looks like you forgot to take the cube root altogether. Removing that I get the same answer.

15. Originally Posted by Reckoner
Are you sure you put it in right? My 89 gave me the correct (approximate) answer. Perhaps you forgot the parentheses around the $\frac13$ or the $3x$?
It's like I always say .... A calculator is very, very stupid and must be treated as such ..... It always does exactly what you tell it to do; but this may not always be what you want it to do.

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