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I'm rather new to Mathematica so there might be an easy solution but I couldn't find it in another forum. I have an integral that will probably require numerical approximation as Mathematica can't solve it directly.

My Code is

xFunction[x_] = (.01 + .3x)/(2*(2 + 3.2*x - 2.39*(x^2)))
$Assumptions = x <= 0.7
$Assumptions = x >= 0.3
MyIntegral = Integrate[(Tanh[Y]^2)*(xFunction[x] - Y)^(1/6), {Y, .3, x}]

Do I need to define x as a real number? The end goal is to take the values of x on 0.3<x<.7 and graph it which x as the independent variable on the x-axis and MyIntegral as the dependent variable on the y-axis. Y is just a dummy variable so it shouldn't matter. I'm sure it's just a few lines of code but I can't seem to figure it out. Any help would be appreciated!

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    $\begingroup$ Please include the full definition of Cfunction and C. $\endgroup$
    – MarcoB
    Commented Jan 28, 2022 at 3:09
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    $\begingroup$ It is good practice to avoid using single capital letters as variable names in Mathematica (in particular, avoid C, since it has special meaning in Wolfram Language). Changing C to c and defining a test polynomial CFunction[x_] := 3 x^2 - 5 x^3, I do get a numerical result: With[{c = 0.6}, NIntegrate[(Tanh[Y]^2)*(CFunction[Y] - Y)^(1/6), {Y, .3, c}]] $\endgroup$
    – ConservedCharge
    Commented Jan 28, 2022 at 4:25
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    $\begingroup$ Do you mean (xFunction[x] - Y) in the integrand and not (xFunction - Y)? $\endgroup$
    – Michael E2
    Commented Jan 28, 2022 at 6:19
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    $\begingroup$ It has nothing to do with x. The integral Integrate[(Tanh[Y]^2)*(Y)^(1/6), Y] does not evaluate analytically. Need to use Numerical integration as shown above. $\endgroup$
    – Nasser
    Commented Jan 28, 2022 at 8:09

1 Answer 1

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By this point, you have done most of the work already. Use NIntegrate instead of Integrate for numerical integration. I also changed = to := in the definition of xFunction, and lastly, defined MyIntegral to be another function:

xFunction[x_] := (.01 + .3 x)/(2*(2 + 3.2*x - 2.39*(x^2)))
MyIntegral[x_] := NIntegrate[(Tanh[Y]^2)*(xFunction[x] - Y)^(1/6), {Y, .3, x}]

Now, this produces complex numbers for $x\in [0.3,0.7]$. So, instead of Plot, I use ReImPlot:

ReImPlot[MyIntegral[x], {x, 0.3, 0.7}]

and obtainenter image description here

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