I have a given function:

sol = t D[ri t/(v - b) - (1/v - 1/(v + b)) a/(b Sqrt[t]), t] // FullSimplify

I now need to integrate this function across $v$ within a specific region v1, v2 (which means a relatively 'simple' definite integral of $\int\frac{1}{v}dv$ which would be a $\log_e\frac{v2}{v1}$ kind of solution.

Integrate[sol, {v, v1, v2}]

However try as I might, it seems that MMA cannot quickly/easily solve this integral. My assumption is, that this has problems in the symbolic solution because of (un)specific constraints, so to reduce the work, I tried.

Integrate[sol,{v,v1,v2}, Assumptions -> {v \[Element] Reals, v1 > 0, v2 > 0}]

However this can't be solved either and try as I can to google or check the documentation I can't seem to find how one would get to the correct solution. For relatively 'simple' equations like this, I can easily solve this by hand...or well 'memory' However I have ever increasingly difficult systems to work with and I would like to calculate these kind of integrals.

How does one do definite integrals of 1/x functions to get the expected symbolic solution of Log[x2/x1]?


The problems are related to the fact that the integral has several parameters. Depending on their values the integral may exist, or not.

Let us see.

This is your function

sol=(ri t)/(-b + v) + a/(Sqrt[t] (2 b v + 2 v^2));

Let us solve first an undetermined integral, calculating the terms of your function separately:

Integrate[#, v, Assumptions -> {ri > 0, t > 0, b > 0}] & /@ sol

(*  ri t Log[-b + v] + (a (Log[v]/(2 b) - Log[b + v]/(2 b)))/Sqrt[t]   *)

Now, one can see that in order for the integral to exist one should require v-b>0 v+b>0 and t!=0. Besides, it is useful to define that v1>0 and v2>v1. Of course, if the integration along the real axis corresponds to your intention. With this

 Integrate[sol, {v, v1, v2}, 
  Assumptions -> {ri > 0, t > 0, b > 0, v1 > 0, v2 > v1, v1 > b}]]

in 0.6s yields the result:

(*  {0.619135, 
 ri t Log[(b - v2)/(b - v1)] + (a Log[((b + v1) v2)/(v1 (b + v2))])/(
  2 b Sqrt[t])}  *)

Have fun!


You can immediately get an answer to the indefinite integral:

Integrate[sol, v]

(* Out: (a Log[v] + 2 b ri t^(3/2) Log[-b + v] - a Log[b + v])/(2 b Sqrt[t]) *)

The definite integral churns for a long time without success without assumptions, as you saw. However, if you add assumptions that all parameters are real, and that $v2>v1>0$, you can get an answer as a ConditionalExpression:

  sol, {v, v1, v2}, 
  Assumptions -> {{t, ri, b, a} ∈ Reals, v2 > v1 > 0}

(* Out: 
  ri t Log[(b - v2)/(b - v1)] + (a Log[((b + v1) v2)/(v1 (b + v2))])/(2 b Sqrt[t]),
  b < v1 && b + v1 > 0

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