Equation 75 · Part 2 · No Particle Without a Cosigner

Symbol e^x

\int_0^\infty x(e^x-1)^{-1}dx=\pi^2/6

e^x

What this part means

$e^x$ is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

Its job in the formula

$e^x$ is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

Full expressionSymbol e^xArticle meaning

The passage around this formula

Write the normalized accelerated spectrum in terms of the dimensionless ratio u $\equiv$ c $\Omega$ /a , using the exact Bose-Einstein shape derived above. Carrying out the normalization integral, $\int_0^\infty$ $\Omega$ \,( $e^{2\pi c\Omega/a}$ -1)^{-1}\,d $\Omega$ = $a^2$ /(24 $c^2$ ) , using the standard result $\int_0^\infty$ x( $e^x$ -1)^{-1}dx= $\pi^2$ /6 , gives the scale-family form

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the article section

[9] Asymptotic States of Accelerated Detectors and Universality of the Unruh Effect ↗

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