# Solving an integral

Given the dimensionless Hubble function:

$$E^2(z) = \biggl ( \frac{H(z)}{H_0} \biggr)^2 = \Omega_m(1+z)^3+ (1-\Omega_m)e^{\int_0^z\frac{1+w0+w1*z}{1+z} dz}$$

I would like to calculate the luminosity distance $$d_l= \frac{(1+z)}{H_0}\int_0^z \frac{H_0}{H(z)}dz$$

as function of $$\Omega_m, w_0,w_1,z$$. For those interested in the meaning of these parameters, the first the matter component of the Universe, the second and the third one are parameters used to describe dark energy and the last one is the redshift which is a positive number.

The integral at the exponential is easy to solve with Assumptions

fun1[y] =
E^Integrate[(1 + w0 + w1*z)/(1 + z), {z, 0, y},
Assumptions ->
z > 0 && z \[Epsilon] Reals && y > 0 && y \[Epsilon] Reals]


and gives

w1 y + (1 + w0 - w1) Log[1 + y]

but the second one is giving me some trouble since Mathematica only gives the symbolic output. I tried with

e[y_] := Sqrt[ omegam *(1 + z)^3 + (1 - omegam)*E^fun1[y]]

Integrate[1/e[z], {z, 0, y}, Assumptions -> z > 0 && z \[Epsilon] Reals && y > 0 && y \[Epsilon] Reals]

but honestly I am not sure anymore since it has been a while since I used Mathematica consistently. I would like to know if I am doing something wrong.

• What 's known about the parameters w0,w1,omegam ? Nov 30, 2022 at 8:37
• @UlrichNeumann ah yes i forgot to write that H_0 is 73 km/(s*parsec) Nov 30, 2022 at 8:55

Try with additional assumptions w0,w1,omegam

Integrate[1/Sqrt[omegam*(1 + z)^3 + (1 - omegam)*E^fun1[z]] , {z, 0, y},
Assumptions ->omegam > 0 && w0 == -1 && w1 == 0 && z > 0 && z \[Epsilon] Reals &&y > 0 && y \[Epsilon] Reals]


Integrate gives the result depending on some extra conditions.

Hope it helps!

• in the previous comment i explained poorly the issue. we don't really know if the lcdm model is correct so i cannot really place those conditions on w0 and w1. all we know is that since the lcdm explain a lot of things the real values should be closer to it. i can add the assumption on omega_m tho. Nov 30, 2022 at 9:30
• In your ?deleted? comment you recommended w1==0&&w0==-1, that's what i tried in my answer Nov 30, 2022 at 10:32
• i know i know, but after you posted your answer i thought about it and realized i was wrong. Nov 30, 2022 at 15:41