How can I get v(t) analytically?

[madhan ravi]

doc dsolve

Answer 1

[Walter Roberson]

I think it unlikely that you can get that analytically. In my experience when integrals involve exp() of square root of a variable, then the integrals rarely seem to have an analytic form.

Furthermore, when a function appears as the limit of an integral that defines the derivative of the function, that is a difficult situation to deal with analytically except for very simple cases.

I notice that the two integrals have the same body, with the multiplication factor k just having been moved from before the exp() in the first to after the exp() in the second, which is not something that has any effect on the value at all (algebraically equivalent.) When I look and see the lower and upper limits being used, my thought runs towards "Cauchy Principle Value". However if you look closely, the order of the integration for the first one is dp_x dk and for the second one is dk dp_x so the where for the first term the integral from 0 to infinity is over p_x, for the second term, the integral from 0 to infinity is over k. That makes the system more difficult to analyze.

I think there is no analytic solution to the equation given. I also suspect that the equation might be wrong: there is no good reason to have moved the k factor or to have exchange the order of the variables for the integration. Anyone trying for clarity would use exactly the same body and same order of integration but would have just changed the limits as needed.

[mohamed elkasemy]

Yes, there is some mistake in writing equation first. Here the right form .But how can solve analytically with that integral.

[Walter Roberson]

Try the dsolve that madhan suggested . If it does not accidentally work then I would recommend giving up.