Math Help - Partial fractions and a funky integral

1. Partial fractions and a funky integral

QUESTION #1:

Suppose we have a partial fraction as follows:

$\frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}$

Can we break that up into partial fractions like so:

$\frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}$

Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?

QUESTION #2:

How in the world would I go about integrating the following function?

$\int\frac{x^5}{e^x-1}dx$

Thanks!

2. Originally Posted by hatsoff
QUESTION #1:

Suppose we have a partial fraction as follows:

$\frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}$

Can we break that up into partial fractions like so:

$\frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}$

Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?
yes

QUESTION #2:

How in the world would I go about integrating the following function?

$\int\frac{x^5}{e^x-1}dx$

Thanks!
i doubt this integral has a closed-form solution. you can wait to see if Krizalid comes up with anything. or maybe Mathstud can show you how to use series to solve it

3. Hello,

Originally Posted by hatsoff
QUESTION #1:

Suppose we have a partial fraction as follows:

$\frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}$

Can we break that up into partial fractions like so:

$\frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}$

Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?
Yes you can
As soon as the degree of the numerator doesn't exceed the denominator's one ^^

4. The integral makes sense only if $x\in[0,\infty).$

Actually, the integral can be tackled by using series.

5. Originally Posted by Jhevon
yes
Neato.

i doubt this integral has a closed-form solution. you can wait to see if Krizalid comes up with anything. or maybe Mathstud can show you how to use series to solve it
It's actually the indefinite form of a definite integral taken from one of Mathstud's problems--but I didn't want to spam his thread by asking high-school-level questions.

What do you mean "closed-form solution"?

$I=\int \frac{x^5}{e^x-1} dx$

$x \rightarrow -x$

$I=\int \frac{x^5}{e^{-x}-1} dx$

Multiply above and below by $-e^x$ :

$I=\int \frac{-x^5 e^x}{e^x-1} dx$

-------------

$2I=\int \frac{x^5}{e^x-1} dx+\int \frac{-x^5 e^x}{e^x-1} dx=\int \frac{x^5 (1-e^x)}{e^x-1}=-\int x^5 dx=-\frac{x^6}{6}$

Therefore, $I=-\frac{x^6}{12}$

Is there any mistake ? o.O

7. Originally Posted by Krizalid
Actually, the integral can be tackled by using series.
yes, i saw that afterwards...

8. Originally Posted by Moo

Is there any mistake ? o.O
Originally Posted by Moo

$I=\int \frac{x^5}{e^x-1} dx$

$x \rightarrow -x$
You can't do that, its application only works for definite integrals.

9. I see ! This is definitely a problem

10. Originally Posted by Moo
I see ! This is definitely a problem
It was a valiant effort though

11. Originally Posted by hatsoff
QUESTION #1:

Suppose we have a partial fraction as follows:

$\frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}$

Can we break that up into partial fractions like so:

$\frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}$

Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?

QUESTION #2:

How in the world would I go about integrating the following function?

$\int\frac{x^5}{e^x-1}dx$

Thanks!
Two things

I have never seen this indefinite integral before but..

$\int_0^{\infty}\frac{u^{x-1}}{e^x-1}dx=\Gamma(x)\zeta(x)=(x-1)!\sum_{n=1}^{\infty}\frac{1}{n^x}$

alternatively

$\int\frac{x^5}{e^x-1}dx=\int\frac{x^4}{1+\frac{x}{2!}+\frac{x^2}{3!}+ ...\frac{x^{n-1}}{n!}}dx$

Choosing the accuracy at which you wish you can tuncate the bottom to a doable amount of terms

12. Originally Posted by Mathstud28
Two things

I have never seen this indefinite integral before but..

$\int_0^{\infty}\frac{u^{x-1}}{e^x-1}dx=\Gamma(x)\zeta(x)=(x-1)!\sum_{n=1}^{\infty}\frac{1}{n^x}$

alternatively

$\int\frac{x^5}{e^x-1}dx=\int\frac{x^4}{1+\frac{x}{2!}+\frac{x^2}{3!}+ ...\frac{x^{n-1}}{n!}}dx$

Choosing the accuracy at which you wish you can tuncate the bottom to a doable amount of terms
Also I have seen this done with complex analysis...might want to wait this one out for TPH or someone more knowledgable in that field

I have seen this split into two integrals...and you turn the second one into a Laurent series

13. We can write this in the form of a polylogarithm. You have probably seen the series. $\sum_{n=1}^{\infty}\frac{z^{n}}{n^{a}}$

The first term is what Moo came up with:

$\frac{-1}{6}x^{6}+x^{5}ln(1-e^{x})+\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{2} }\cdot{5x^{4}}$

$-\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{3}}\cdot{ 20x^{3}}+\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{ 4}}\cdot{60x^{2}}$

$-\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{5}}\cdot{ 120x}+\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{6}} \cdot{120}$

See the pattern?.

14. Originally Posted by Moo

$I=\int \frac{x^5}{e^x-1} dx$

$x \rightarrow -x$

$I=\int \frac{x^5}{e^{-x}-1} dx$

Multiply above and below by $-e^x$ :

$I=\int \frac{-x^5 e^x}{e^x-1} dx$

-------------

$2I=\int \frac{x^5}{e^x-1} dx+\int \frac{-x^5 e^x}{e^x-1} dx=\int \frac{x^5 (1-e^x)}{e^x-1}=-\int x^5 dx=-\frac{x^6}{6}$

Therefore, $I=-\frac{x^6}{12}$

Is there any mistake ? o.O

I did a very similar mistake in an integral posted in mathlinks recently. And a nice reader enlightened me with this example as to why it is wrong:

Originally Posted by kevinatcausa
The problem's the same thing that's wrong with the following argument.

$\int 2x^3 dx = \int (x^2)( 2x dx) = \int u du = \int x dx = \frac {x^2}{2} + c$

The trouble is that $\int u du = \frac {u^2}{2} + c$, while $\int x dx = \frac {x^2}{2} + c$. They may look the same, but they're not equal because $u$ and $x$ are different.

15. Originally Posted by Isomorphism

I did a very similar mistake in an integral posted in mathlinks recently. And a nice reader enlightened me with this example as to why it is wrong:
Yes, because when changing variable, either you change the boundaries, either you substitute back at the end. That's why you can't do this way