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∫0∞x(ex−1)−1dx=π2/6\int_0^\infty x(e^x-1)^{-1}dx=\pi^2/6

Why this formula appears here

Write the normalized accelerated spectrum in terms of the dimensionless ratio u ≡\equiv cΩ\Omega/a , using the exact Bose-Einstein shape derived above. Carrying out the normalization integral, ∫0∞\int_0^\infty Ω\Omega\,(e2πcΩ/ae^{2\pi c\Omega/a}-1)^{-1}\,dΩ\Omega = a2a^2/(24c2c^2) , using the standard result ∫0∞\int_0^\infty x(exe^x-1)^{-1}dx=π2\pi^2/6 , gives the scale-family form

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exe^x

Symbol e^x

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

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00

Starting index or lower bound: 0

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∞\infty

Ending index or upper bound: infty

This label says where the repeated addition, multiplication, or accumulation stops. It sets the last term or end of the range.

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Published contexts (1)

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∫0∞x(ex−1)−1dx=π2/6\int_0^\infty x(e^x-1)^{-1}dx=\pi^2/6

Equation 75 · Evolutionary Physics

No Particle Without a Cosigner

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

Write the normalized accelerated spectrum in terms of the dimensionless ratio u ≡\equiv cΩ\Omega/a , using the exact Bose-Einstein shape derived above. Carrying out the normalization integral, ∫0∞\int_0^\infty Ω\Omega\,(e2πcΩ/ae^{2\pi c\Omega/a}-1)^{-1}\,dΩ\Omega = a2a^2/(24c2c^2) , using the standard result ∫0∞\int_0^\infty x(exe^x-1)^{-1}dx=π2\pi^2/6 , gives the scale-family form

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