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I have following the following integral,

$$\int_{-1}^{1} \sqrt{(1-a+0.5(1+2b))\left(\frac{1}{r^{6}-1+i 0^{+}}- (a+2b)\right)}dr= \Pi$$

I want to draw a plot of how $b$ vary when $a$ varies from $0.5$ to $1$. I can't evaluate this integral symbolically due to the singularity. To remove the singularity I can add a small imaginary part to the integration.

$$\int_{-1}^{1} \sqrt{(1-a+0.5(1+2b))\left(\frac{1}{r^{6}-1+i 0^{+}}- (a+2b)\right)}dr= \Pi$$

But even after that, I have the difficulty of finding a way to numerically solve this equation in Mathematica to find $a$ and $b$ values.

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1 Answer 1

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Maybe so:

f[b_?NumericQ, a_?NumericQ] := With[{ep = 10^-12(*You may change it*)}, 
NIntegrate[Sqrt[(1 - a + 1/2*(1 + 2 b))*(1/(r^6 - (1 + I*ep)) - (a + 2*b))], {r, -1, 1}, 
Method -> "LocalAdaptive"]]

Π = 1/10;(*You may change it*)
ContourPlot[Re[f[b, a]] == Π, {a, 1/2, 1}, {b, -1.2, -0.4},
FrameLabel -> {"a", "b(a)"}, PlotPoints -> 50]

enter image description here

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  • $\begingroup$ Thank you this is what I wanted :) $\endgroup$
    – Coderzz
    Aug 4, 2017 at 1:47

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