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Here's the integral I have,

$$ \displaystyle\int\dfrac{x^4}{1+ e^x} dx $$

I tried the usual methods I know, but I failed miserably.

How would you all approach this problem?

Martin Sleziak
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William
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    Why do you want to calculate this integral? I suspect that you actually have another problem which involves this integral. – md2perpe Jul 20 '18 at 19:58
  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Jul 20 '18 at 20:00
  • @md2perpe Not really I'm just playing around with integrals. (Even though I'm a rookie) – William Jul 20 '18 at 20:01
  • The integral is very similar to the one here. – md2perpe Jul 20 '18 at 20:01
  • @md2perpe Yea I know, I did check that a while back, but it wasn't much help :( – William Jul 20 '18 at 20:03
  • Not all elementary functions have a primitive function that is an elementary function. This is probably such a case. – md2perpe Jul 20 '18 at 20:07
  • @md2perpe Trust me, everytime I thought that an integral wont have an elementary solution, someone came and surprised me. So I can't say for sure. – William Jul 20 '18 at 20:09
  • RayDansh has given you probably the best answer you can get for the indefinite integral, but if you'd only wanted the $\int_0^\infty dx$ value it would've been easier, viz. $\int_0^\infty\frac{x^4 e^{-x}dx}{1+e^{-x}}=24\eta(5)=\frac{45}{2}\zeta(5)$. – J.G. Jul 20 '18 at 20:18
  • @J.G. ikr, that's why I said, it's hard cuz it's indefinite :( – William Jul 20 '18 at 20:20
  • may be he mean $\displaystyle \int^{a}_{-a}\frac{x^4}{1+e^x}dx,a>0$ – DXT Jul 23 '18 at 06:12

2 Answers2

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According to WolframAlpha, you would have to use nonelementary functions such as the polylogarithm function, $\operatorname{Li}_n(x)$: $$\frac{x^5}{5} - x^4 \log(1 + e^x) - 4 x^3 \operatorname{Li}_2(-e^x) + 12 x^2 \operatorname{Li}_3(-e^x) - 24 x \operatorname{Li}_4(-e^x) + 24 \operatorname{Li}_5(-e^x)$$

J.G.
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RayDansh
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$\int\frac {x^4}{1+e^x} \ dx\\ \int\frac {x^4e^{-x}}{1+e^{-x}} \ dx$

Converting to a geometric series: $\frac {y}{1+y} = \sum_\limits{n=1}^\infty (-y)^n$

$\int x^4\sum_\limits{n=1}^\infty (-1)^ne^{-nx} \ dx$

considering just one term $\int x^4e^{-nx} = (\frac {x^4}{n} + \frac {4x^3}{n^2} + \frac {12x^2}{n^2} + \frac {24x}{n^3} + \frac {24}{n^4}) e^{-nx}$

$\sum_\limits{n=1}^{\infty}(-1)^n(\frac {x^4}{n} + \frac {4x^3}{n^2} + \frac {12x^2}{n^2} + \frac {24x}{n^3} + \frac {24}{n^4}) e^{-nx}$

It is worth noting $\frac{1}{\Gamma(s)} \int_0^{\infty} \frac {x^{s-1}}{e^x -1} \ dx = \zeta(s)$

Doug M
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