Markov spectrum

2013-05-19T16:15:38Z

2A01:E35:2F45:9A0:223:14FF:FE3E:4490: Fix Freiman constant, see Weisstein. Beware of the typo in Cusick/Flahive

{{Unreferenced|date=August 2008}}
{{Cleanup|date=December 2008}}

In [[mathematics]], '''[[integral|integration]] by parametric derivatives''' is a method of integrating certain functions.

For example, suppose we want to find the integral

: <math>\int_0^\infty x^2 e^{-3x} \, dx.</math>

Since this is a product of two functions that are simple to integrate separately, repeated [[integration by parts]] is certainly one way to evaluate it. However, we may also evaluate this by starting with a simpler integral and an added parameter, which in this case is ''t'' = 3:

: <math>
\begin{align}
& \int_0^\infty e^{-tx} \, dx = \left[ \frac{e^{-tx}}{-t} \right]_0^\infty = \left( \lim_{x \to \infty} \frac{e^{-tx}}{-t} \right) - \left( \frac{e^{-t0}}{-t} \right) \\
& = 0 - \left( \frac{1}{-t} \right) = \frac{1}{t}.
\end{align}
</math>

This converges only for ''t'' > 0, which is true of the desired integral. Now that we know

: <math>\int_0^\infty e^{-tx} \, dx = \frac{1}{t},</math>

we can differentiate both sides twice with respect to ''t'' (not ''x'') in order to add the factor of ''x''<sup>2</sup> in the original integral.

: <math>
\begin{align}
& \frac{d^2}{dt^2} \int_0^\infty e^{-tx} \, dx = \frac{d^2}{dt^2} \frac{1}{t} \\[10pt]
& \int_0^\infty \frac{d^2}{dt^2} e^{-tx} \, dx = \frac{d^2}{dt^2} \frac{1}{t} \\[10pt]
& \int_0^\infty \frac{d}{dt} \left (-x e^{-tx}\right) \, dx = \frac{d}{dt} \left(-\frac{1}{t^2}\right) \\[10pt]
& \int_0^\infty x^2 e^{-tx} \, dx = \frac{2}{t^3}.
\end{align}
</math>

This is the same form as the desired integral, where ''t'' = 3. Substituting that into the above equation gives the value:

: <math>\int_0^\infty x^2 e^{-3x} \, dx = \frac{2}{3^3} = \frac{2}{27}.</math>

[[Category:Integral calculus]]

{{mathanalysis-stub}}

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