Bandwidth of an energy signal I'm studying for my midterm , and I'm struggling with finding bandwidths. 

I have this energy spectral density : $\Psi(f)= { 1  \over \sqrt {  1 +  ({f \over B }) ^2} }$ ,  and I need to find  the $3$ $dB$ band and the band with $90%$ % power. 

About the $90%$ % case, do I need to find the total energy and then  $\int_0^W \Psi(f) = 0.9 \times  Total Energy$?  

One think that doesn't make sense for me is that I'm dealing with an energy signal but I need to find something related to power.

If I had a power signal and wanted to find the 3 dB bandwidth, would I need to find the average power and then solve $\int_0^W S_x(f) = 0.5 \times TotalPower$ ?

Thanks in advance, and sorry for the bad english.
I'm using the following assumption: the magnitude of the spectrum is given by the formula

$$|X(f)|=\frac{1}{\sqrt{1+(f/B)^2}}$$

Then the energy spectral density is

$$|X(f)|^2=\frac{1}{1+(f/B)^2}\tag{1}$$

The 3 dB bandwidth is simply the frequency for which (1) becomes $0.5$, which is $W_{3dB}=B$. The bandwidth according to the 90% energy/power definition is

$$\int_{-W}^W|X(f)|^2df=0.9\cdot\int_{-\infty}^{\infty}|X(f)|^2df\tag{2}$$

Equation (2) is the reason why I assume the definition as given in (1), because otherwise the integral on the right-hand side of (2) does not converge, so the 90% energy/power definition would not make sense. With the definition (1) the integrals in (2) are easily solved using the $\arctan$ function, from which $W=const\cdot B$ follows, where $const=6.31$ (if I'm not mistaken), but I'm sure you can do the calculation yourself.