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I am having hard time to figure out if there is a way to get a close-form of the following integral: $$\int\limits_0^{ + \infty } {{{\left( {\frac{1}{{{\sigma ^2}}}} \right)}^{N + 3}}\exp \left\{ { - \frac{1}{{2{\sigma ^2}}}{Y^T}\left[ {I - {R^{ - 1}}} \right]Y + \frac{1}{\sigma }{Y^T}{R^{ - 1}}\xi } \right\}d} \sigma % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbeqaaeaacaGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % qadaqaamaalaaabaGaaGymaaqaaiabeo8aZnaaCaaaleqabaGaaGOm % aaaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaad6eacqGHRaWkca % aIZaaaaOGaciyzaiaacIhacaGGWbWaaiWaaeaacqGHsisldaWcaaqa % aiaaigdaaeaacaaIYaGaeq4Wdm3aaWbaaSqabeaacaaIYaaaaaaaki % aadMfadaahaaWcbeqaaiaadsfaaaGcdaWadaqaaiaadMeacqGHsisl % caWGsbWaaWbaaSqabeaacqGHsislcaaIXaaaaaGccaGLBbGaayzxaa % GaamywaiabgUcaRmaalaaabaGaaGymaaqaaiabeo8aZbaacaWGzbWa % aWbaaSqabeaacaWGubaaaOGaamOuamaaCaaaleqabaGaeyOeI0IaaG % ymaaaakiabe67a4bGaay5Eaiaaw2haaiaadsgaaSqaaiaaicdaaeaa % cqGHRaWkcqGHEisPa0Gaey4kIipakiabeo8aZbaa!630E! $$

R : N*N SDP Matrix.

I: Identity Matrix.

Y : N*1 vector .

ξ : N*1 vector.

Thank you for your answers and guidance.

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  • $\begingroup$ Avoid using constants with a complex expression, they just add clutter. $\endgroup$ – Yves Daoust Feb 26 at 10:27
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If I am right, with $u:=\dfrac1\sigma$, the integral is of the form

$$\int_0^\infty u^{n+1}e^{-au^2/2+bu}du.$$

Then by a linear change of variable and assuming $a>0$, you turn it to

$$\int_0^\infty\left(v-c\right)^{n+1}e^{-v^2/2}dv.$$

By the Binomial theorem, this is a sum of the $k^{th}$ order moments of the Gaussian, for which formulas are available (or can be retrieved by integration by parts).


Update:

My bad, after the change of variable, the lower bound is no more $0$, so that the integrals require the incomplete Beta function (or the Error function). So there is no hope of a closed-form without special functions.

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  • $\begingroup$ Thank you for your answer Yves Daoust I made that variable change and get a similar formula but the lower side of the integral is not zero but $$\frac{b}{{\sqrt a }} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbeqaaeaacaGaaiaabaqaamaabaabaaGcbaWaaSaaaeaaca % WGIbaabaWaaOaaaeaacaWGHbaaleqaaaaaaaa!37EA! $$ which make thing more complicated $\endgroup$ – Ayoub Sbitti Feb 26 at 10:58
  • $\begingroup$ @AyoubSbitti: ooops, right. $\endgroup$ – Yves Daoust Feb 26 at 11:02
  • $\begingroup$ My original problem arise from a probabilistic calculation of posterior probability distribution of hyperparameters.this hyperparameters have to be integrated or maximized to get de MAP (Maximum a Posteriori) of that hyperparameters $\endgroup$ – Ayoub Sbitti Feb 26 at 11:29

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