What is the relation between the power spectral density and the variance of the input time series?

The variance of the series is a time-domain statistic.  Its frequency domain counterpart is the spectral density function PSD.  The PSD is the Fourier transform  of the variance. That's the relationship. 

Probably you are solving a task of noise power estimation at the output of pass-band filter... The answer to your question depends on the noise model you are use. Usually at low frequencies the uniform PSD is used, but at upper frequencies - so called 1/f model (or -3 dB/octave). For example, for OpAmps for low frequency noises we can see one level of PSD, but for high frequencies (usually 1 kHz)  another level which is less (1/f) at high frequencies. Therefore, first of all you are to know what is the law of the input signal PSD, and only than perform integration procedure. May be you will have to divide the integration interval into 2 intervals which will correspond appropriate signal PSD model.

The variance is the power of the variable part of the signal. If you remove the mean of the signal, the power of the remaining signal is the variance. Now, compute the PSD of this signal. Ideally, it s the Fourier transform of the autocorrelation function. The integral of the PSD is equal to te power and equal to the autocorrelation at the origin. These facts are using to "calibrate" the PSD estimates, since the estimation methods do not guarantee the equality of both (time and spectral) estimates of the power.

The problem is caused by non-uniqueness of presentation of the PSD, let it be of some noise signal. Pure sense of the notion  Power Spectral Density is the derivative  of the power   P[f₀,f]  within  an interval of frequencies  [f₀, f]  with respect to  f, i.e.

PDS(f) =  dP[f₀,f] / df

On the other hand, according to Igor's remarks,  there is a custom to present the density  o fthe power recalculated "per  an octave", or some another count of frequencies  with respect to the logaritmic scale. For example, if  the  interval  is presented in the form [f₀=2^x₀ , f = 2^x], then the cumulative power within the interval of length counted in number of octaves equals  PO[x₀,x] = P[f₀,f]. Therefore its density, again expressed by the derivative - this time with respect to  x by the chain rule equals:

PDSO(x) = dPO[x₀,x] / dx = PDS(2^x) * df/dx = PDS(2^x) * 2^x * log(2).

Thus   using  f=2^x  we obtain

 PDS(f)  = PDSO(log₂f ) / [ f * log(2)]

Summarizing, the operation to be performed depends on the given  FORM  of ththe power desnsity. If this is PDSO  [in units W/octave], then one should divide it by the frequency and multiply by the band width. If this is already PDS [in units W/Hz] then for the power of the band it sufficies simply to multiply by the band width.

Of course, from the question we don't know the form of the data:) Therefore it is not possible to give more concrete answer.

Best regards, Joachim