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From: Peter Pein on 13 Jun 2010 04:11
There is no real-valued solution; to get complex solutions, try to give
a complex value as starting point for FindRoot:
target[x_?NumericQ] := NIntegrate[BesselI[0, x*t], {t, 0, 1}] and
FindRoot[target[x]==1/2,{x,1+I}]
{x->0.+2.84653 I}

give what you want?


Am Sat, 12 Jun 2010
09:30:44 +0000 (UTC) schrieb Jason Quinn <jason.lee.quinn(a)gmail.com>:

> Suppose I have an expression of the form
>
> 1/2 = int_0^1 besselI[0,x*t] dt
>
> and I want to find the value of "x" that will make the integral true.
> Can Mathematica handle such situations? I've tried all the main
> suspects that I get warnings that the expression depends on x in a
> non- algebraic way. The numerical tasks do not seem to work either.
> The expression above is just a made-up example, but in general, what
> do to when you are trying to solve equations of this type? I could
> certainly write a function that find the value iteratively myself, but
> I'd be surprised if such a thing doesn't already exist.
>
> Cheers,
> Jason
>
> PS I've tried Wolfram Alpha and
>
> solve 1 = int besselI0(x*t) from t=0 to 1
>
> fails but
>
> solve 1/2 = int erf(x*t) from t=0 to 1
>
> works.
>



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