Interesting integral here:

$$\int_0^\infty \frac{\Gamma(D/2+\nu/2)}{\Gamma(\nu/2)}\frac{|\Lambda|^{1/2}}{(\pi \nu)^{D/2}}\left(1+\frac{(x-\mu)^{T}\Lambda(x-\mu)}{\nu}\right)^{-D/2-\nu/2}\mathrm dx$$

I tried to write $\Lambda$ as $\Lambda^{1/2}\Lambda^{1/2}$ and changing variables with $y =\Lambda^{1/2}(x-\mu)$ in order to get:

$$\int_0^\infty \frac{\Gamma(D/2+\nu/2)}{\Gamma(\nu/2)}\frac{|\Lambda|^{1/2}}{(\pi \nu)^{D/2}}\left(1+\frac{y^{T}y}{\nu}\right)^{-D/2-\nu/2}|\Lambda|^{-1/2}\mathrm dy_{1}\mathrm dy_{2}\cdots \mathrm dy_{n}$$

The integral in the form:

$$\int_0^\infty \left(1+\frac{y^{2}_{1}+y^{2}_{2}+...+y^{2}_{n})}{\nu}\right)^{-D/2-\nu/2}\mathrm dy_{1}\mathrm dy_{2} \cdots \mathrm dy_{n}$$

seems to generate a product of Gamma function but I can't get a closed form.

Any help is appreciated. By the way, this question is somewhat related to Simplifying covariance matrices in distributions, so maybe one can help the other.

  • $\begingroup$ why would that integral not diverge horribly? $\endgroup$
    – Jonathan
    Commented Mar 9, 2013 at 0:52
  • $\begingroup$ Because of the power −D/2−ν/2 (which I forgot to add in the third equation.) Fixed now. $\endgroup$
    – r_31415
    Commented Mar 9, 2013 at 1:01

1 Answer 1


As you have noticed, the integral can be transformed into $$\int_0^\infty d^ny \frac{1}{(1+y^Ty)^\alpha}.$$ Going to the spherical coordinates leads to $$\frac{1}{2^n}\int d\Omega_n\int_0^\infty dr \frac{r^{n-1}}{(1+r^2)^\alpha},$$ where the factor of $1/2^n$ is due to integrating over this fraction of the whole space and $\int d\Omega_{n}$ is the volume of $S^{n-1}$, which is equal to $$\int d\Omega_{n} = \frac{2 \pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)},$$ as can be shown for example by evaluating the integral $\int d^ny e^{-y^Ty}$ both in Cartesian and spherical coordinates and comparing the two results.

To evaluate the simple integral over $r$, make the change of variables $x=(1+r^2)^{-1}$, which leads to $$\int_0^\infty dr \frac{r^{n-1}}{(1+r^2)^\alpha} = \frac{1}{2}\int_0^1 x^{\alpha-n/2-1}(1-x)^{n/2-1}dx=\frac{\Gamma(\alpha-n/2)\Gamma(n/2)}{2\Gamma(\alpha)},$$ where the expression for the Beta function was used in the last step. Putting the pieces together, you arrive at $$\int_0^\infty d^ny \frac{1}{(1+y^Ty)^\alpha} = \left(\frac{\pi}{4}\right)^{n/2}\frac{\Gamma(\alpha-n/2)}{\Gamma(\alpha)}.$$

  • $\begingroup$ Thanks! I'm still not sure about the $\frac{1}{2^{n}}$. Can you provide a reference? $\endgroup$
    – r_31415
    Commented Mar 9, 2013 at 2:22
  • $\begingroup$ There would be no $\frac{1}{2^n}$ if your integrals were $\int_{-\infty}^{\infty}$. Because you're only integrating over positive values of $y$s, you have to cut down by half for each dimension. $\endgroup$
    – D M
    Commented Mar 9, 2013 at 2:41
  • $\begingroup$ Oh, I understand. Great answer. $\endgroup$
    – r_31415
    Commented Mar 9, 2013 at 2:50
  • $\begingroup$ I think this integral should be from $-\infty$ to $\infty$ because it's supposed to be normalized and following the procedure you described (including $\nu$) everything gets cancelled out except the $1/2^{n}$. $\endgroup$
    – r_31415
    Commented Mar 9, 2013 at 3:53

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