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Find $a\in \mathbb{R}$ such that $$\int_{-a}^a \frac{x^4}{e^x+1}dx = -\frac{32}{5}$$

Kamal Saleh
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Raluca lok
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  • Use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 OR https://math.stackexchange.com/questions/1073120/integral-int-12011-frac-sqrtx-sqrt2012-x-sqrtxdx – lab bhattacharjee Dec 11 '19 at 13:59

2 Answers2

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Note,

$$\int_{-a}^a \frac{x^4}{e^x+1}dx = \int_{-a}^0 \frac{x^4}{e^x+1}dx +\int_{0}^a \frac{x^4}{e^x+1}dx $$ $$= \int_{0}^a \frac{t^4e^t}{e^t+1}dt +\int_{0}^a \frac{x^4}{e^x+1}dx = \int_0^a x^4dx=\frac15a^5 = -\frac{32}{5}$$

where $t=-x$ is used in the first integral. Thus, $a = -2$.

Quanto
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  • How can the integral of a non-negative function over any interval be negative? – Adrian Keister Dec 11 '19 at 15:19
  • @AdrianKeister - Due to the reversed limits. – Quanto Dec 11 '19 at 15:21
  • Oh, gotcha. When someone writes an integral like $\int_{-a}^a f(x),dx,$ the natural thing is to assume $a>0.$ So the real problem is that no one should ever write an integral like that unless they DO mean $a>0.$ Otherwise, we have a failure of rhetoric. – Adrian Keister Dec 11 '19 at 15:26
  • @ViktorGlombik - Appreciate the suggestion. I'd consider integrals like this somewhat specialized due to certain symmetry, which WA's algorithms could not recognize yet. – Quanto Dec 11 '19 at 20:14
  • @KamalSaleh That is correct. I deleted that comment since I realized that the proposed alternative is actually very very close to the solution presented here. – ViktorStein Jul 25 '23 at 21:12
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You are looking for a real number $a$ such that

$$ \int_{-a}^a \frac{x^4}{e^x+1}dx= - \frac{32}{5} ?$$

No number with this property exists, since $\frac{x^4}{e^x+1} \ge 0$ for all $x$ and therefore

$$\int_{-a}^a \frac{x^4}{e^x+1}dx >0.$$

Fred
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