# Please evaluate this double integral over rectangular bounds

• {???}
In summary: Thanks,QMIn summary, the integral in question is a difficult one that requires iterated integration, and the integrand is correct despite its unusual dimensions. The indefinite integral has been derived and the integration constants cancel when using rectangular bounds. The definite integral still remains to be evaluated.
{???}
Summary:: Could someone please evaluate this double integral over rectangular bounds? Answer only is just fine.

[Mentor Note -- thread moved from the technical math forums, so no Homework template is shown]

Hi,

I'm trying to find the answer to the following integral over the rectangle ##(x^-,x^+)\times(y^-,y^+)##:
$$\int_{y^-}^{y^+}\int_{x^-}^{x^+}v^2\arctan\frac{Lu}{v\sqrt{u^2+v^2+L^2}}\,\mathrm{d}v\,\mathrm{d}u.$$
For the indefinite double integral, WA gives a complicated answer with logarithms of complex arguments. I know that ##\tan^{-1}z=\frac{i}{2}\log\frac{1-iz}{1+iz}##, but it is not clear how to get this from the WA expression:
$$\log\left\{-\frac{12(L\sqrt{L^2+u^2+v^2}+L^2+u^2-iuv)}{L^2u^3(v+iu)}\right\}$$
Obviously, this should have a real answer since the integrand is real-valued over the domain of integration.

I don't have access to Mathematica at the moment and the internet integral evaluators I've tried all seem to fail here. I'm not interested in the step-by-step approach (##u##-/trigonometric substitution, integration by parts, partial fractions, etc.). I just want to know the answer to the double integral above.

Can somebody with access to a CAS or Mathematica please compute this double integral over the given rectangular bounds and share the answer? It would be much appreciated.

Thanks,
QM

Last edited by a moderator:
WWGD and Delta2
##u(x^-,x^+)## or ##v(x^-,x^+)## ?

Where is Origin ? You do not care ?

{?} said:
Summary:: Could someone please evaluate this double integral over rectangular bounds? Answer only is just fine.

Hi,

I'm trying to find the answer to the following integral over the rectangle ##(x^-,x^+)\times(y^-,y^+)##:
$$\int_{y^-}^{y^+}\int_{x^-}^{x^+}v^2\arctan\frac{Lu}{v\sqrt{u^2+v^2+L^2}}\,\mathrm{d}v\,\mathrm{d}u.$$
For the indefinite double integral, WA gives a complicated answer with logarithms of complex arguments. I know that ##\tan^{-1}z=\frac{i}{2}\log\frac{1-iz}{1+iz}##, but it is not clear how to get this from the WA expression:
$$\log\left\{-\frac{12(L\sqrt{L^2+u^2+v^2}+L^2+u^2-iuv)}{L^2u^3(v+iu)}\right\}$$
Obviously, this should have a real answer since the integrand is real-valued over the domain of integration.

I don't have access to Mathematica at the moment and the internet integral evaluators I've tried all seem to fail here. I'm not interested in the step-by-step approach (##u##-/trigonometric substitution, integration by parts, partial fractions, etc.). I just want to know the answer to the double integral above.

Can somebody with access to a CAS or Mathematica please compute this double integral over the given rectangular bounds and share the answer? It would be much appreciated.

Thanks,
QM
What is the application? Is this for schoolwork?

Sorry that was a typo. Should be ##u(x^-,x^+)##.

berkeman said:
What is the application? Is this for schoolwork?
It is a schoolwork-type problem, but not schoolwork. I was investigating electrostatic properties of an analytically soluble charge distribution when I encountered these integrals.

{?} said:
It is a schoolwork-type problem, but not schoolwork.
Okay, please post schoolwork-type questions in the schoolwork forums, and please show as much work on the question as you can. I will move your thread to the schoolwork forums for you now.

This thread will help to explain the PF rules about schoolwork-type questions:

{?} said:
Summary:: Could someone please evaluate this double integral over rectangular bounds? Answer only is just fine.

[Mentor Note -- thread moved from the technical math forums, so no Homework template is shown]

Hi,

I'm trying to find the answer to the following integral over the rectangle ##(x^-,x^+)\times(y^-,y^+)##:
$$\int_{y^-}^{y^+}\int_{x^-}^{x^+}v^2\arctan\frac{Lu}{v\sqrt{u^2+v^2+L^2}}\,\mathrm{d}v\,\mathrm{d}u.$$
Mathematica is giving a complex answer because without explicitly defining the domain of L, x, and y, Mathematica assumes they are complex. To restrict L, a, and b to Reals, use the code:
Mathematica:
ClearAll[theL, a, b];
Integrate[
v^2 ArcTan[(theL u)/(v Sqrt[u^2 + v^2 + theL^2])], {v, -b,
b}, {u, -a, a}, Assumptions -> Element[{theL, a, b}, Reals]]

Your integral smells fishy to me. I assume we have
$$\frac{y}{x}=\tan(\phi)$$
$$\frac{L}{\sqrt{x^2+y^2 +L^2}}=\cos(\theta)$$
where ##\phi## is the azimuthal angle and ##\theta## is the angle of elevation measured from the vertical. Your integrand is
$$x^2\tan^{-1}(\tan(\phi)\cos(\theta))$$
with the dimensions of ##angle*(length)^2##. That's just weird for an electrostatics problem with rectangular symmetry. It looks like your mixing apples and oranges. Where did you get that? It should be a red flag for you that something went wrong in your derivation when WA gives a complex solution.

Last edited:
Dimensions of ##angle\cdot length^2## are correct. There are other factors of ##x_{\pm}^{-1},y_{\pm}^{-1},z_{\pm}^{-1},L^{-1}## and these lead to dimensions of ##[k]charge^2length^{-1}## when all is said and done. I just omitted those prefactors to focus on the integral I have not yet been able to evaluate. I am absolutely certain the integrand is correct, it's just an uncommon (and hard) problem requiring enough iterated integration that this integral eventually occurs.

I have worked out the indefinite integral:
$$\iint v^2\tan^{-1}\frac{Lu}{v\sqrt{L^2+u^2+v^2}}\,\mathrm{d}u\,\mathrm{d}v =2Lu(u^2+L^2)\sinh^{-1}\frac{v}{\sqrt{u^2+L^2}}-\tfrac{1}{2}L^4\tan^{-1}\frac{uv}{L\sqrt{u^2+v^2+L^2}}$$
$$-\tfrac{1}{2}u^4\tan^{-1}\frac{Lv}{u\sqrt{u^2+v^2+L^2}} +\tfrac{1}{2}v^4\tan^{-1}\frac{Lu}{v\sqrt{u^2+v^2+L^2}} +\tfrac{5}{2}Luv\sqrt{u^2+v^2+L^2} +u^3v\sinh^{-1}\frac{L}{\sqrt{v^2+u^2}},$$
Integration constants are omitted because they cancel when I input rectangular bounds.

Can someone please confirm this result?

If it were single integral the definite integral should be
$$[RHS]_{x-}^{x+}$$
I am not sure how we should calculate definite double integral from RHS. May I do it as
$$[[RHS]_{x-}^{x+}]_{y-}^{y+}$$?

For confirmation how about calculating and checking
$$\frac{\partial RHS}{\partial u\partial v}=Integrand$$?

Integrand is odd function for both u and v. So we expect RHS is even function for both u and v like
$$\int dx \int dy \ \ xy= \frac{x^2y^2}{4}$$
But your RHS is odd function for both u and v. Is my speculation wrong ?

Last edited:

## 1. What is a double integral?

A double integral is a type of mathematical operation that involves integrating a function of two variables over a two-dimensional region. It is represented by two integral signs and is used to calculate the volume under a surface in three-dimensional space.

## 2. What are rectangular bounds?

Rectangular bounds refer to the limits or boundaries of the region over which the double integral is being evaluated. In other words, it defines the area that is being integrated over in the x-y plane.

## 3. How do I evaluate a double integral over rectangular bounds?

To evaluate a double integral over rectangular bounds, you first need to determine the limits of integration for both variables, typically denoted by a and b for x and c and d for y. Then, you can use the basic rules of integration to solve the integral, which involves finding the antiderivative of the function and plugging in the limits of integration.

## 4. What is the purpose of evaluating a double integral over rectangular bounds?

Evaluating a double integral over rectangular bounds allows you to find the volume under a surface in three-dimensional space. It is used in many fields of science, such as physics, engineering, and economics, to solve problems involving areas and volumes.

## 5. Are there any special cases when evaluating a double integral over rectangular bounds?

Yes, there are some special cases when evaluating a double integral over rectangular bounds. For example, if the function being integrated is symmetric about one of the axes, the integral can be simplified by using the symmetry to reduce the number of integrals. Additionally, if the function is separable, the integral can be written as a product of two single integrals.

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