Plotting the solution of an integral equation

Posted 3 months ago
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 Hello All, I tried plotting the solution of the integral equations in this notebook, but the plot function gives empty plot. The solution contains Hypergeometric functions. Please guide what needs correction. Thanks.
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Posted 3 months ago
 Your second integral is poorly typed. You almost surely need to insert a space between c and e. I wonder if c and C are meant to be different parameters. Then you set up an equation but you don't solve it. The equation does not depend on x, but you try to plot it as a function of x, which does not make much sense. Also, before plotting I suppose you need to give numerical values to most parameters.
Posted 3 months ago
 This is the updated notebook with numerical values for the coefficients(and yes, c and C are two different parameters). Now, the notebook contains only one integral equation whose solution needs to be plotted for the value of k=2.5. Thanks for your response.
Posted 3 months ago
 You what plot of T vs x, but where is x in integral solution ?There is only T.
Posted 3 months ago
 Your equation does not depend on x, only on T: diff = Subtract @@ Simplify[sol, T > 0]; Plot[diff, {T, 0, 2}] FindRoot[diff == 0, {T, 1.7}] There is a value of T that makes the two integrals equal.
Posted 3 months ago
 Alright. so, the plot should be over a wide range of values of T, but when i change the range in the Plot command from say {T, 0, 10}, the values go infinite. i have attached the updated notebook in which i have tried to plot T from 0 to 10. And the attached image shows how the actual plot should look like(here i have used k=2.5, so blue curve is relevant).
Posted 3 months ago
 A = 0.3136; B = 0.0692; c0 = 0.6207; a = 1.29; b = 0.1101; c = 3.014; sol[T_, k_] := NIntegrate[x (1 + x/((k - 3/2) T))^(-k - 1), {x, 0, Infinity}] - NIntegrate[ x (1 + x/((k - 3/2) T))^(-k - 1) (c*E^(-(x/a)) - c*E^(-x/b) - B*E^(-c0*x) + A), {x, 0, Infinity}] Plot[sol[T, 2.5], {T, 0, 3}](*for: k=2.5*) FindRoot[sol[T, 2.5], {T, 2},WorkingPrecision -> 20] (*{T ->1.7115048830163596308}*) The plot is different because it is not normalized.
Posted 3 months ago
 Oh okay. So, how can I normalize it?
Posted 3 months ago
 I don't know what functions are plotted in the book. However, in your own plot you need to increase the precision because there are numerical instabilities for larger values of T: diffExact = Rationalize[diff, 0]; Plot[diffExact, {T, 0, 10}, WorkingPrecision -> 30, PlotRange -> {-2, 1.5}]