# Integration - Engineer in need!

• Apr 27th 2008, 09:10 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!!
• Apr 27th 2008, 09:14 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)
• Apr 27th 2008, 09:24 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. :/
• Apr 27th 2008, 09:28 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
• Apr 27th 2008, 09:33 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)

• Apr 27th 2008, 10:34 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.
• Apr 27th 2008, 10:58 AM
CaptainBlack
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
• Apr 27th 2008, 11:27 AM
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.
• Apr 27th 2008, 11:29 AM
john5115
Ok, so what do you do with these polylogarithm things?

Thanks
• Apr 27th 2008, 12:18 PM
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}$
• Apr 27th 2008, 01:40 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
• Apr 27th 2008, 01:45 PM
CaptainBlack
Quote:

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
• Apr 27th 2008, 01:48 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
• Apr 27th 2008, 02:22 PM
john5115
well, ol' Captain Black, i think you are over estimating my mathematical ability!

Still, i'll give it ago!