Australia Telescope National Facility

The HI Mass Function from HIPASS
V. Kilborn ,
R. L. Webster ,
L. Staveley-Smith ,
, PASA, 16 (1), in press.

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Results

**Figure 3:** Mass distribution for the flux limited sample.
$\begin{figure} \begin{center} \centerline{\psfig{file=kilborn3.epsi,height=10cm,width=12cm,angle=-90}} \end{center} \end{figure}$

The standard

$\sum 1 / V_{\rm max}$ method (Schmidt 1968) was used to determine the HIMF from the data. A Schechter function (Schechter 1976) was then fitted to the data:

$\begin{displaymath} \Phi\left({M_{HI} \over M^*}\right)d\left({M_{HI} \over M^*... ...{M_{HI} \over M^*}\right) d\left({M_{HI}\over M^*}\right), \end{displaymath}$

(1)

where M^* is the mass that defines the characteristic knee in the function, and $\theta^*$ is the normalization. Figure 4 shows the HIMF derived using this method, and the best fit Schechter function is shown by the solid line. The slope of the low-mass end is $\alpha = 1.32$ , the characteristic mass, $\log M^*=9.5$ , and the normalization, $\theta ^* = 0.01$ . These are similar to values found by Zwaan et al. (1997), who find $\alpha = 1.2$ , and logM^* = 9.55 (see the dotted line in Figure 4). The normalization found for the HIPASS sample is slightly lower than the Zwaan et al. value of 0.014. There is no indication from our data that there is a turn-up in the slope of the HIMF to the mass limit we have imposed. However, as we still have poor statistics in the low mass bins (a single galaxy in the lowest mass bin), the faint end slope is still somewhat uncertain. A maximum likelihood method (Sandage et al. 1979) was also used to determine Schechter parameters for the data. The Schechter parameters determined using this method are slightly different from the best fit parameters found using the

$\sum 1 / V_{\rm max}$ method, with

$\alpha = 1.15$ and

$\log M^* = 9.74$ . Figure 5 shows a plot of contours of likelihood for the sample. It is possible that the difference in value of the

$\sum 1 / V_{\rm max}$ method and the maximum likelihood method is due to clustering of the galaxies in this region. The maximum likelihood method is not dependent on the distribution of galaxies in the sample, whereas the

$\sum 1 / V_{max}$ varies according to the clustering of the galaxies in the sample (Binggeli et al. 1988). A full investigation of selection effects and different computation methods will be presented in a later paper. Using the

$\sum 1 / V_{max}$ Schechter function parameters derived for this sample we can determine the HI density at the present epoch, which is

$\rho_{HI} = \Gamma(2 - \alpha)M^* \theta^*$ , where $\Gamma$ is the Euler gamma function. We obtain

$\rho_{HI}= 4.2 \times 10^7 hM_{\odot}$ Mpc^-3, or

2.8 x 10^-33h g cm^-3. The cosmological mass density of HI at z = 0, $\Omega_{HI}$ , is the ratio of the HI density over the critical density at the present epoch (

$\rho_c=1.88 \times 10^{-29}h^2$ g cm^-3 Padmanabhan, 1993). This gives a value of

$\Omega_{HI} = 1.5 \times 10^{-4} h^{-1}$ . Assuming that the mass percentage of He I is 25%, the total cosmological mass density of neutral gas at the present epoch is

$\Omega_g = 1.88 \times 10^{-4}h^{-1}$ : this is slighty lower than the Zwaan et al. (1997) value of

$\Omega_g(z=0)=(2.7 \pm 0.5) \times 10^{-4}h^{-1}$ .

**Figure 4:** HI mass function (HIMF) - The solid points represent the measured HI mass function per decade, and the solid line represents the best fit Schechter function for the
$\sum 1 / V_{\rm max}$ method from this sample: $\alpha = 1.32$ ,

$\log M^*=9.5$ and

$\theta ^* = 0.01$ . The dotted line is the HIMF from Zwaan et al. (1997), with $\alpha = 1.2$ and

$\log M^* = 9.55$ . The error bars show $1 \sigma$ errors (Marshall 1985).
$\begin{figure} \begin{center} \centerline{\psfig{file=kilborn4.epsi,height=10cm,width=10.5cm,angle=-90}} \end{center} \end{figure}$

**Figure 5:** Contours from the maximum likelihood method. The maximum value is seen at $\alpha = 1.15$ ,
$\log M^* = 9.74$ . The contours are 68.3% and 99.9% confidence limits.
$\begin{figure} \begin{center} \centerline{\psfig{file=kilborn5.epsi,height=9cm,angle=180}} \end{center} \end{figure}$

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Contents Page: Volume 16, Number 1

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