In Wolfram Alpha I computed the following double integral:

enter image description here

What does $i$ mean here? Does the result have imaginary part also?

  • 2
    $\begingroup$ Yes, it denotes the imaginary unit. $\endgroup$
    – Gary
    Dec 12, 2022 at 11:49

2 Answers 2


It does mean imaginary, but notice: the imaginary part of the answer printed by Wolfram is only $$4.54647\times10^{-13}$$ which is another way to write $$0.000000000000454647.$$

What probably happened here was: Wolfram used some advanced technique to evaluate the integral, which involved complex numbers. The imaginary parts of its calculation should have canceled out to zero. But computer calculations aren't always perfectly accurate, and in this case there were some small errors that remained at the end.

So the imaginary part is likely just an error, and the correct answer has imaginary part equal to zero as you would expect.


Hoping that you want to know why an imaginary part (even very, very small appears.

When the inner integral is computed, it is in fact the result of $$I=\int_{-20}^{20} z\, \tanh ^{-1}\left(\frac{\sqrt{(z-A-1)\, (z+A-1)}}{B}\right)\,dz $$ where $$A=\sqrt{-y^2+20 y-299} \qquad \text{and} \qquad B=y-10$$ that is to say $$A=\color{red}{\mathbf{i}} \sqrt{\left(y-\big[10-\color{red}{\mathbf{i}} \sqrt{199} \big]\right)\,\left(y-\big[10+\color{red}{\mathbf{i}} \sqrt{199} \big]\right)}$$

I shall not type here the result of the integration; one of the terms (randomly selected) is $$\coth ^{-1}\left(\frac{B}{\sqrt{361-A^2}}\right)=\coth ^{-1}\left(\frac{y-10}{\sqrt{y^2-20 y+660}}\right)$$ whose antiderivative is $$-\sqrt{y^2-20 y+660}+10 \log \left(\sqrt{y^2-20 y+660}-y+10\right)+y \coth ^{-1}\left(\frac{y-10}{\sqrt{y^2-20 y+660}}\right)$$ Computing its value for $y=20$ gives $$-2 \sqrt{165}+10 \log \left(2 \sqrt{165}-10\right)+20\color{red}{\coth ^{-1}\left(\sqrt{\frac{5}{33}}\right)}$$

$$\coth ^{-1}\left(\sqrt{\frac{5}{33}}\right)=\frac 12 \log \left(\frac{\sqrt{165} +33}{\sqrt{165} -33}\right)=\frac 12 \left(\color{red}{\mathbf{i}}\pi +\log \left(\frac{19+\sqrt{165}}{14} \right)\right)$$

There are many situations like that but, at the end, all complex numbers combine to give excatly $0$.

I have in front of me the exact result of this definite integral; there is no complex number. Only the numerical evaluation could lead to a number close to the machine precision.


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