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This integral is needed to obtain the joint distribution of the sample variances of a random sample from a bivariate Gaussian distribution. For details on the joint distribution of the sample means, sample variances, and sample correlation, please see Sections 14.11 to 14.13 of Kendall's book (2nd edition):

Kendall, Maurice G., The Advanced Theory of Statistics. Vol. I, Philadelphia: J. B. Lippincott Co., XI, 457 p. (1944). ZBL0063.03214.

The integral is $$ I\left( n,c\right) =\int_{-1}^{1}\exp\left( cr\right) \left( 1-r^{2}\right) ^{\left( n-4\right) /2}dr,$$

where $c\neq0$ is a constant, and $n\geq4$ is (a constant and) an integer (and is the sample size). When $n-4$ is even, we can use ``integration by parts'' to get $I\left( n,c\right) $ after some tedious computations. In general, we can use the expansion $$\left( 1-r^{2}\right) ^{\left( n-4\right) /2}=\sum_{k=0}^{\infty}a_{k,n}r^{2k},$$ compute each $$s_{k}=\int_{-1}^{1}\exp\left( cr\right) a_{k,n}r^{2k}dr,$$ and then compute $\sum_{k=0}^{\infty}s_{k}$. For this method, we may not have an explicit, analytic expression for $I\left( n,c\right) $.

On the other hand, we can use the expansion $$\exp\left( cr\right) =\sum_{k=0}^{\infty}\frac{c^{k}r^{k}}{k!},$$ compute $$b_{2k}=2\int_{0}^{1}% \frac{c^{2k}}{\left( 2k\right) !}r^{2k}\left( 1-r^{2}\right) ^{\left( n-4\right) /2}dr,$$ and compute $\sum_{k=0}^{\infty}b_{2k}$. Note that each $b_{2k}$ relates to the Beta function (upon a change of variable $r^{2}% \mapsto\tau$). But what does $\sum_{k=0}^{\infty}b_{2k}$ look like?

Could someone please point me to some references on whether $I\left( n,c\right) $ has an explicit, analytic expression, or on whether such an expression can be obtained by one of the attempts mentioned above? Dose this integral involve special functions? Thanks.

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  • $\begingroup$ Personally I don't think it's appropriate to post this kind of question without including results from Wolfram Alpha/Mathematica/Maple/Sage. $\endgroup$ – Neil Strickland Nov 22 at 9:54
  • $\begingroup$ Thank you, Neil. Will do that. $\endgroup$ – Chee Nov 23 at 1:12
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It seems that Wolfram Alpha understands this integral. Its answers are given in terms of modified Bessel functions of the first kind $I_k$.

It seems that the result is, for $A(k,c):=I(n,c)$, with $n-4=2k+1$,

$$I(n,c)=\frac{(2k+1)!!\cdot \pi\cdot I_{k+1}(c)}{c^{k+1}}.$$

For instance, $$\begin{eqnarray}I(5,c)&=&A(0,c)=\frac{\pi I_{1}(c)}{c}\\ I(7,c)&=&A(1,c)=\frac{3\pi I_{2}(c)}{c^2}\\ I(9,c)&=&A(2,c)=\frac{5\cdot 3\cdot \pi\cdot I_{3}(c)}{c^3}\\ I(11,c)&=&A(3,c)=\frac{7\cdot 5\cdot 3\cdot \pi\cdot I_{4}(c)}{c^4} \end{eqnarray}$$ The last equation for $I(11,c)$ is verified by Wolfram Alpha here.

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  • $\begingroup$ Kjos-Hanssen: thank you for pointing out the Bessel function. I think I can get it going by using contour integral by raising the $cos$ function in the definition of the Bessel function to some rational power . $\endgroup$ – Chee Nov 22 at 6:06
  • $\begingroup$ Prof. Kjos-Hanssen: how did you get the formular? Thank you. $\endgroup$ – Chee Nov 22 at 6:16
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    $\begingroup$ @Chee just educated guessing based on the Wolfram Alpha answers... $\endgroup$ – Bjørn Kjos-Hanssen Nov 22 at 6:28
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Using the integral representation of the modified Bessel function $I$ (see DLMF, for example), $$ I_{\nu}\left(z\right)=\frac{(\frac{1}{2}z)^{\nu}}{\pi^{\frac{1}{2}}\Gamma\left(\nu+\frac{1}{2}\right)}\int_{-1}^{1}(1-t^{2})^{\nu-\frac{1}{2}}e^{\pm zt}\mathrm{d}t $$ we find directly $$I(n,c)=2^{\frac{n-3}{2}}\sqrt{\pi}\Gamma\left( \frac{n}{2}-1 \right)c^{\frac{3-n}{2}}I_{\frac{n-3}{2}}(c)$$

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  • $\begingroup$ Paul Enta: thank you for pointing this out, saving me time on computing a contour integral on my own. Neil Strickland: i am most of the time a statistician, and do not often use Mathematica :) $\endgroup$ – Chee Nov 23 at 20:27

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