Integration - Engineer in need!

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April 27th 2008, 08:10 AM
john5115

Integration - Engineer in need!

Hello people; here's how it goes, i'm writing my masters dissertation at the moment about the mechanical properties of metallic glasses (interesting stuff eh?!!) and i've got to integrate a function. Now, i got 34% on my undergrad calc test so, try as i might, i just can't do it! Please can someone throw me a lifeline!!

Here's the function to integrate y = (x^4.e^x)/((e^x-1)^2) dx

Thanks!!
April 27th 2008, 08:14 AM
Moo

Hello,

You can do an integration by parts :

$u(x)=x^4$

$v'(x)=\frac{e^x}{(e^x-1)^2}$ , which is something like $\frac{g'(x)}{g^2(x)}$

You should know an antiderivative for it :)

And then, iterate the integration by parts 3 times (I think)
April 27th 2008, 08:24 AM
flyingsquirrel

Hi

Quote:

Originally Posted by Moo

And then, iterate the integration by parts 3 times (I think)

That's what I thought too but when you get to $\int \frac{4x^3}{e^x-1}\mathrm{d}x$ I'm not sure you can go on. :/
April 27th 2008, 08:28 AM
Mathstud28

Quote:

Originally Posted by flyingsquirrel

Hi

That's what I thought too but when you get to $\int \frac{4x^3}{e^x-1}\mathrm{d}x$ I'm not sure you can go on. :/

Yeah you cannot I am pretty sure
April 27th 2008, 08:33 AM
Moo

Quote:

Originally Posted by flyingsquirrel

Hi

That's what I thought too but when you get to $\int \frac{4x^3}{e^x-1}\mathrm{d}x$ I'm not sure you can go on. :/

(Swear)

Too bad ^^
April 27th 2008, 09:34 AM
Krizalid

It can't be solved, but by takin' $\int_0^\infty\frac{x^3}{e^x-1}\,dx$ one can solve it, by using $\zeta(4)$ result.
April 27th 2008, 09:58 AM
CaptainBlack

1 Attachment(s)

Quote:

Originally Posted by john5115

Hello people; here's how it goes, i'm writing my masters dissertation at the moment about the mechanical properties of metallic glasses (interesting stuff eh?!!) and i've got to integrate a function. Now, i got 34% on my undergrad calc test so, try as i might, i just can't do it! Please can someone throw me a lifeline!!

Here's the function to integrate y = (x^4.e^x)/((e^x-1)^2) dx

Thanks!!

You are not an undergraduate now, this is not for a calculus course. So you do what a professional would do, use technology or look it up.

Quick maths gives the attachment ( $Li_n$ is the polylogarithm)

RonL
April 27th 2008, 10:27 AM
Danshader

Quote:

Originally Posted by CaptainBlack

You are not an undergraduate now, this is not for a calculus course. So you do what a professional would do, use technology or look it up.

Quick maths gives the attachment ( $Li_n$ is the polylogarithm)

RonL

i am sorry if i am being impolite but could you show how the calculations would go if i don't using technology to solve it.
April 27th 2008, 10:29 AM
john5115

Ok, so what do you do with these polylogarithm things?

Thanks
April 27th 2008, 11:18 AM
PaulRS

But you can find a beautiful formula for the improper Integral.

Note that: $ \int_0^\infty {\frac{{x^{s - 1} \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \int_0^\infty {\frac{{\tfrac{{x^{s - 1} }} {{e^x }}}} {{\left( {1 - \tfrac{1} {{e^x }}} \right)^2 }}dx} = \int_0^\infty {x^{s - 1} \sum\limits_{n = 1}^\infty {n \cdot e^{ - xn} } \cdot dx} $

$ \int_0^\infty {\frac{{x^{s - 1} \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \sum\limits_{n = 1}^\infty {n \cdot \int_0^\infty {x^{s - 1} \cdot e^{ - xn} \cdot dx} } $

Now: $ \int_0^\infty {x^{s - 1} \cdot e^{ - xn} \cdot dx} = \frac{{\Gamma \left( s \right)}} {{n^s }} $

Thus: $ \int_0^\infty {\frac{{x^{s - 1} \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \Gamma \left( s \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{1} {{n^{s - 1} }}} } \right) = \Gamma \left( s \right) \cdot \zeta \left( {s - 1} \right) $

In particular: $ \int_0^\infty {\frac{{x^3 \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \Gamma \left( 4 \right) \cdot \zeta \left( 3 \right)$

And: $ \int_0^\infty {\frac{{x^4 \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} =\Gamma \left( 5 \right) \cdot \zeta \left( 4 \right)=\frac{4\cdot{\pi^4}}{15}$
April 27th 2008, 12:40 PM
CaptainBlack

Quote:

Originally Posted by PaulRS

But you can find a beautiful formula for the improper Integral.

Note that: $ \int_0^\infty {\frac{{x^{s - 1} \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \int_0^\infty {\frac{{\tfrac{{x^{s - 1} }} {{e^x }}}} {{\left( {1 - \tfrac{1} {{e^x }}} \right)^2 }}dx} = \int_0^\infty {x^{s - 1} \sum\limits_{n = 1}^\infty {n \cdot e^{ - xn} } \cdot dx} $

$ \int_0^\infty {\frac{{x^{s - 1} \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \sum\limits_{n = 1}^\infty {n \cdot \int_0^\infty {x^{s - 1} \cdot e^{ - xn} \cdot dx} } $

Now: $ \int_0^\infty {x^{s - 1} \cdot e^{ - xn} \cdot dx} = \frac{{\Gamma \left( s \right)}} {{n^s }} $

Thus: $ \int_0^\infty {\frac{{x^{s - 1} \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \Gamma \left( s \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{1} {{n^{s - 1} }}} } \right) = \Gamma \left( s \right) \cdot \zeta \left( {s - 1} \right) $

In particular: $ \int_0^\infty {\frac{{x^3 \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} = \Gamma \left( 4 \right) \cdot \zeta \left( 3 \right)$

And: $ \int_0^\infty {\frac{{x^4 \cdot e^x }} {{\left( {e^x - 1} \right)^2 }}dx} =\Gamma \left( 5 \right) \cdot \zeta \left( 4 \right)=\frac{4\cdot{\pi^4}}{15}$

But we are not asked for definite integrals. If we are doing engineering we don't worry about this hard work we just compute a numerical integral.

RonL
April 27th 2008, 12:45 PM
CaptainBlack

Quote:

Originally Posted by Danshader

i am sorry if i am being impolite but could you show how the calculations would go if i don't using technology to solve it.

Look up polylogarithm on wikipedia or mathworld, and you will see how they might occur when this thing is integrated by parts.

RonL
April 27th 2008, 12:48 PM
CaptainBlack

Quote:

Originally Posted by john5115

Ok, so what do you do with these polylogarithm things?

Thanks

The same thing that you would do with anyother special function, look them up, use asymtotic forms to investigate behaviour for large arguments, use series to approximate for small arguments, ...

RonL
April 27th 2008, 01:22 PM
john5115

well, ol' Captain Black, i think you are over estimating my mathematical ability!

Still, i'll give it ago!