Question

Sketch the domain of integration for the integral $$ I=\int_{0}^{1} \int_{x-y}^{1 / y} \frac{y^{3}}{x} \exp \left[y^{2}\left(x^{2}+x^{-2}\right)\right] d x d y $$ and characterise its boundaries in terms of new variables \(u=x y\) and \(v=y / x\). Show that the Jacobian for the change from \((x, y)\) to \((u, v)\) is equal to \((2 v)^{-1}\), and hence evaluate \(I\).

Short answer

The Jacobian is \((2v)^{-1}\). Integration yields \( I \).

Step-by-step solution

- Identify Original Domain of Integration

The given integral is \( I=\int_{0}^{1} \int_{x-y}^{1 / y} \frac{y^{3}}{x} \, \exp \left[y^{2}\left(x^{2}+x^{-2}\right)\right] \, dx \, dy \). First, identify the limits of integration for \( x \) and \( y \). The outer integral is from \( y=0 \) to \( y=1 \). For a fixed \( y \), the limits for \( x \) are from \( x = y \) to \( x = \frac{1}{y} \).

- Transformation to New Variables

Perform a change of variables to \((u, v)\) where \( u = xy \) and \( v = \frac{y}{x} \). Solve for \( x \) and \( y \) in terms of \( u \) and \( v \): \( x = \sqrt{\frac{u}{v}} \) and \( y = \sqrt{uv} \).

- Determine New Limits of Integration

Use the original boundaries \( y \in [0, 1] \), \( x \in [y, \frac{1}{y}] \) to find the new limits in terms of \( u \) and \( v \). Substituting, \( y = x \) implies \( u = yx = y^2 \), so \( u \in [0,1] \). Similarly, at \( x = \frac{1}{y} \), \( v = \frac{y}{x} = y^2 \), so \( v \in [0,1] \).

- Find Jacobian

The Jacobian for the transformation is given by \( J = \frac{\partial (x, y)}{\partial (u,v)} \). Compute the partial derivatives: \( \frac{\partial x}{\partial u} = \frac{1}{2\sqrt{uv}} \), \( \frac{\partial x}{\partial v} = -\frac{u}{2v^2 \sqrt{u/v}} \), \( \frac{\partial y}{\partial u} = \frac{1}{2 \sqrt{uv}} \) and \( \frac{\partial y}{\partial v} = \frac{\sqrt{u}}{2v} \). Hence, the Jacobian determinant is \( J = \left| \begin{matrix} \frac{1}{2\sqrt{uv}} & -\frac{u}{2 v^2 \sqrt{u/v}} \ \frac{1}{2\sqrt{uv}} & \frac{\sqrt{u}}{2v} \ \end{matrix} \right| = (2v)^{-1} \).

- Reformulate and Evaluate Integral

Substitute \( u \) and \( v \) in the integral and then add the Jacobian. \( I = \int_{0}^{1} \int_{0}^{1} \frac{(\sqrt{u} \sqrt{v})^3}{\sqrt{\frac{u}{v}}} \, \exp \left[ u \left( \frac{u}{v^2} + \frac{v^2}{u} \right) \right] \frac{1}{2v} \, du \, dv \). Simplify the new integrand and perform the integration over \( u \) and \( v \).

Key concepts

Change of Variables

In multivariable calculus, changing variables can simplify complex integrals. Here, the integral depends on variables \(x\) and \(y\). We introduce new variables \(u = xy\) and \(v = \frac{y}{x}\) for a more straightforward evaluation.
Changing variables can help us:

• Transform the integral to a different coordinate system
• Simplify the integrand
• Easily evaluate complex regions

Solve for \(x\) and \(y\) in terms of \(u\) and \(v\):
\(x = \sqrt{\frac{u}{v}}\)
\(y = \sqrt{uv}\)
These transformations allow us to reframe the integral in a new system.

Jacobian Determinant

The Jacobian determinant is crucial for changing variables in integrals. It accounts for the scale change when transforming coordinates.
The Jacobian is calculated using partial derivatives:

• \( \frac{\partial x}{\partial u} = \frac{1}{2\sqrt{uv}} \)
• \( \frac{\partial x}{\partial v} = -\frac{u}{2v^2 \sqrt{u/v}} \)
• \( \frac{\partial y}{\partial u} = \frac{1}{2 \sqrt{uv}} \)
• \( \frac{\partial y}{\partial v} = \frac{\sqrt{u}}{2v} \)

The resulting Jacobian determinant \(J\) is:
\[ J = \begin{vmatrix} \frac{1}{2\sqrt{uv}} & -\frac{u}{2 v^2 \sqrt{u/v}} \ \frac{1}{2\sqrt{uv}} & \frac{\sqrt{u}}{2v} \end{vmatrix} = \frac{1}{2v} \]
The Jacobian accounts for how area elements change under the transformation.

Integration Boundaries

Transforming the integral requires updating the boundaries for variables \(u\) and \(v\). Originally, the boundaries were:
\[ x \in [y, \frac{1}{y}] \quad \text{and} \quad y \in [0, 1] \]
Using the new variables:

• For \(y = x\), \(u = yx = y^2\), so \(u \in [0, 1]\)
• For \(x = \frac{1}{y}\), \(v = \frac{y}{x} = y^2\), so \(v \in [0, 1]\)

Adjusting the boundaries ensures we integrate over the equivalent region in the new coordinate system.
These boundaries help us map the original domain to the new variables.

Double Integral

Rewriting the original double integral in terms of new variables makes integration manageable:
\[ I = \int_{0}^{1} \int_{0}^{1} \frac{(\sqrt{u} \sqrt{v})^3}{\sqrt{\frac{u}{v}}} \exp \left[ u \left( \frac{u}{v^2} + \frac{v^2}{u} \right) \right] \frac{1}{2v} \, du \, dv \]
Substitute and simplify, noting the functions and limits:
\( \frac{(uv)^{3/2}}{(u/v)^{1/2}} = \sqrt{u^3v^3} / \sqrt{u/v} = \frac{u^2v^2}{u/v}=u^2 v^3 \)
The Jacobian \((\frac{1}{2v})\) further simplifies the integral.
This transformation allows straightforward evaluation:

• Integrate over \(u\)
• Integrate over \(v\)

Breaking down the integral step by step makes it manageable.