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Today, I tried to obtain the solution of the following equation, 

sol = DSolve[ {m v'[t] + \[Gamma] v[t] == \[Xi][t] , v[0] == v0}, v[t], t]

The solution, of course, can be easily obtained and the result was,

$$ \left\{\left\{v(t)\to e^{-\frac{\gamma t}{m}} \left(-\int_1^0 \frac{\xi (K[1]) e^{\frac{\gamma K[1]}{m}}}{m} \, dK[1]+\int_1^t \frac{\xi (K[1]) e^{\frac{\gamma K[1]}{m}}}{m} \, dK[1]+\text{v0}\right)\right\}\right\}$$

This is correct solution but I am surprised that two integrals were present in this solution. Then, I tried to simplify using functions like FullSimplify and so on, but I failed to simplify it. Is there any automatic way to accomplish this goal?

I am using Mathematica 10.0 on the linux machine.

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  • $\begingroup$ to further simplify you need to know something about the range of t. try specifying assumptions. $\endgroup$
    – george2079
    Oct 22, 2014 at 11:51
  • $\begingroup$ @george2079: Thank you. I tried to add some assumptions by defining $Assumptions = {\[Gamma] > 0, m > 0, t > 0, v0 > 0};. But, at least, this approach does not give a simpler form. $\endgroup$
    – Sungmin
    Oct 22, 2014 at 12:02
  • $\begingroup$ mathematically you need 0<=t<=1 to consolidate that to a single integral (from 0 to t). (I don't know if mathematica will do the simplify even under that assumption though) $\endgroup$
    – george2079
    Oct 22, 2014 at 12:38
  • $\begingroup$ @george2079 I am not very sure that your condition 0<= t<= 1 is really needed. Anyway, this additional assumption does not give a simpler solution. Thank you anyway. $\endgroup$
    – Sungmin
    Oct 22, 2014 at 12:46
  • $\begingroup$ right of course, caught me commenting before my second cup of coffee. I'll point out you can show this more simply as Integrate[ f[x] , {x, 0, 1} ] + Integrate[ f[x] , {x, 1, t } ] does not simplify under any assumptions I can think of. $\endgroup$
    – george2079
    Oct 22, 2014 at 14:13

1 Answer 1

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you might handle this with a TransformationFunction:

 Simplify[ 
    Integrate[ f[x] , {x, 0, 1} ] + Integrate[ f[x] , {x, 1, t } ] ]
 t[s_Plus?(
      Length[#] == 2 &&
      Head[#[[1]]] == Integrate &&
      Head[#[[2]]] == Integrate &&
      #[[1, 1]] == #[[2, 1]] &&
      #[[1, 2, 3]] == #[[2, 2, 2]] &)] := 
       Integrate[s[[1, 1]], {s[[1, 2, 1]], s[[1, 2, 2]], s[[2, 2, 3]]}]
 Simplify[ 
    Integrate[ f[x] , {x, 0, 1} ] + Integrate[ f[x] , {x, 1, t } ] , 
    TransformationFunctions -> {Automatic , t }]

enter image description here

I'm sure this can be made cleaner and more robust..for example you need to deal with reversing limits and changing signs, and so on (eg:)

 t[s_Times?(#[[1]] == -1 &&  Head[#[[2]] ] == Integrate &)] :=
    Integrate[s[[2, 1]], {s[[2, 2, 1]], s[[2, 2, 3]], s[[2, 2, 2]]}]
 t[s_Plus?(
        Length[#] == 2 &&
        Head[#[[1]]] == Integrate &&
        Head[#[[2]]] == Integrate &&
        #[[1, 1]] == #[[2, 1]] &&
        #[[1, 2, 3]] == #[[2, 2, 2]] &)] := 
             Integrate[s[[1, 1]], {s[[1, 2, 1]], s[[1, 2, 2]], s[[2, 2, 3]]}]
 Simplify[ -Integrate[ f[x] , {x, 1, 0} ] + Integrate[ f[x] , {x, 1, t } ] , 
         TransformationFunctions -> {Automatic , t }]

same result

For your example this works if you do:

 Simplify[sol /. v0 -> 0, TransformationFunctions -> {Automatic, t }]

to be more general you'd need to coax simplify to test every combination of terms in multi term sum.

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