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There are a lot of integral representations for $\pi$ as well as infinite series, limits, etc. For other transcendental constants as well (like $\gamma$ or $\zeta(3)$).

However, for every definite integral that is equal to $e$ I can think of, the integrated function contains the exponent in some way.

Can you provide some definite integrals that have $e$ as their value (or some elementary function of $e$ that is not a logarithm), without $e$ appearing in any way under the integral or as one of its limits (and without the limits for $e$, or the infinite series for $e$)?

The example or what I want is the following integral for $\pi$:

$$\int_0^{1} \sqrt{1-x^2} dx=\frac{\pi}{4}$$

Yuriy S
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  • Note that your example is just the area under a quarter of a circle! – zz20s Mar 03 '16 at 13:26
  • I assume that hyperbolic functions or trigonometric functions should not be an example? – S.C.B. Mar 03 '16 at 13:27
  • Based on Quadarture rules, Brothers and Knox '98 gave several non-trivial representations for $e$. See http://www.ccsenet.org/journal/index.php/jmr/article/viewFile/3724/3320 –  Mar 03 '16 at 13:30
  • Possibly Related:http://math.stackexchange.com/questions/1653979/is-there-any-integral-for-the-golden-ratio(I'm not suggesting this question is a duplicate-just saying the two questions seem similar. – S.C.B. Mar 03 '16 at 13:31
  • @Travis, thanks for your comment. – NoChance Mar 03 '16 at 13:34
  • Yes, hyperbolic functions are out. Limits as well (except for $\infty$ as the integral limit). Bacon, I see only limits in the paper you offered – Yuriy S Mar 03 '16 at 13:46
  • $e=\sum_{x=0}^{\infty} \frac{1}{1.2.3...(x-3)(x-2)(x-1) \int_{t}^{\sqrt{2x+t^{2}}} p ~ dp}$ – NoChance Mar 03 '16 at 16:22

1 Answers1

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If you are asking whether e is a period, the question remains, technically, still open, but its answer is not expected to be affirmative. As to the non-algebraic integrands, we have $$\int_{-\infty}^\infty\frac{\cos(ax)}{1+x^2}~dx~=~\frac\pi{e^{|a|}}$$

Lucian
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  • Thank you. I forgot about this kind of integrals, since they are usually evaluated by using the complex exponential – Yuriy S Mar 03 '16 at 17:45