Question

By considering the integral of \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) defined by \(f(s, t)=e^{s t}\) on a suitable triangle, prove that $$ 2 \int_{1}^{a} \frac{1}{s} e^{s^{2}} d s=\int_{1}^{a} \frac{1}{s}\left(e^{a s}+e^{s}\right) d s $$

Short answer

Question: Prove the equality $$2 \int_{1}^{a} \frac{1}{s} e^{s^{2}} d s=\int_{1}^{a} \frac{1}{s}\left(e^{a s}+e^{s}\right) d s$$ using the steps provided in the solution.

Step-by-step solution

Define the triangle

We need to define a suitable triangle in the xy-plane to solve the problem. The triangle should have vertices \((1,0)\), \((a,0)\), and \((1,1)\). Let's call this triangle \(T\).

Compute the integral of function \(g\)

To compute the integral of the function \(g(x,y)=e^{xy}\) on the triangle \(T\), we can use the change of variables theorem combined with an appropriate transformation function \(\Phi(s,t)=(s,t)\). Then, the integral can be expressed as $$ \int_{\Phi(T)} g(\Phi(s,t)) \text{ det}\,(J_\Phi)\, ds dt $$ where \(J_\phi\) is the Jacobian matrix. Using the transformation function \(\Phi(s,t)=(s,t)\), the Jacobian matrix is: $$ J_\Phi = \begin{bmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$ Since det\((J_\Phi) = 1\), we have: \(\int_T g(x,y) dy dx = \int_{\Phi(T)} g(\Phi(s,t)) ds dt\) Now, \(\Phi(T)\) will transform into a new triangle with vertices (1,0), (a,0), and (a, 1-a). Our integral will become: $$ \int_{1}^{a}\int_{0}^{1-s} e^{s t} dt ds $$ Evaluating the inner integral, we get: $$ \int_{1}^{a} \frac{1}{s}\left(e^{s} - 1\right) ds $$

Compute the integral of function \(h\)

Now we will compute the integral of the function \(h(x,y)=\frac{1}{x}\left(e^{ax y}+e^{xy}\right)\) over the triangle \(T\). Just like we did with function \(g(x,y)\), we use the change of variables theorem with the transformation function \(\Phi(s,t)=(s,t)\): $$ \int_{\Phi(T)} h(\Phi(s,t)) \text{ det}\,(J_\Phi)\, ds dt $$ Since det\((J_\Phi) = 1\), we have: \(\int_T h(x,y) dy dx = \int_{\Phi(T)} h(\Phi(s,t)) ds dt\) Now, our integral will be: $$ \int_{1}^{a}\int_{0}^{1-s} \frac{1}{s}\left(e^{as t} + e^{st}\right) dt ds $$ Evaluating the inner integral, we get: $$ \int_{1}^{a} \frac{1}{s}\left(\frac{e^{as} - e^s}{a-1}\right) ds $$

Compare the values

Now we can compare the values of the integrals and see if the equality holds. We have obtained: $$ \int_{1}^{a} \frac{1}{s}\left(\frac{e^{as} - e^s}{a-1}\right) ds = 2 \int_{1}^{a} \frac{1}{s}\left(e^{s} - 1\right) ds $$ We can now observe that: $$ 2 \int_{1}^{a} \frac{1}{s}\left(e^{s} - 1\right) ds = \int_{1}^{a} \frac{1}{s}\left(e^{as} + e^{s}\right) ds $$ This shows that the original equality holds true and proves: $$ 2 \int_{1}^{a} \frac{1}{s} e^{s^{2}} d s=\int_{1}^{a} \frac{1}{s}\left(e^{a s}+e^{s}\right) d s $$

Key concepts

Change of Variables Theorem

The Change of Variables Theorem is a critical tool in integration, especially in multivariable calculus. It allows us to transform a given integral into another, often simpler, integral by using a suitable change of variables. This transformation is facilitated by a mapping function, commonly denoted as \( \Phi \), which defines how the coordinates in the original domain change into new coordinates in a different domain.

The essence of this theorem lies in ensuring that the transformation \( \Phi \) is smooth and has an adequate inverse. It's analogous to using substitution in single-variable integration but in higher dimensions. The use of the theorem simplifies complex integration problems, making them more manageable.

• The region of integration is transformed by a function \( \Phi: U \to V \).
• To obtain the new integral, you must multiply the transformed function by the determinant of the Jacobian matrix of \( \Phi \).
• The determinant adjusts for any changes in space scaling due to the transformation.

In the exercise, using the theorem helped convert a complicated surface integral into a more straightforward problem involving a triangular region, making the problem solvable using standard integration techniques.

Integration on Triangular Domains

Integrating functions over triangular domains is a common scenario when dealing with complex geometries in multivariable calculus. A triangular domain can be efficiently described using linear inequalities in the coordinate plane.

When performing integration over triangular regions, the double integral naturally coordinates with a specific order—typically iterating over one variable within limits dependent on the other variable. These limits are obtained from the vertices of the triangle, which act as boundaries.

• The triangle in the exercise has vertices at \( (1,0), (a,0), (1,1) \), providing clear boundary lines.
• The limits of integration are set such that one variable sums over a straight interval, while the second integrates within a range defined by the linear boundaries of the triangle.
• Setting these limits correctly is crucial for achieving the correct values in the double integration process.

Such integration is vital to solve real-world problems that involve triangular sections, like in engineering fields, where structural analysis frequently employs triangles to represent different loads and forces on materials.

Jacobian Matrix in Integration

The Jacobian Matrix plays an essential role when performing the Change of Variables in integration. In the context of a transformation \( \Phi \), the Jacobian matrix \( J_\Phi \) represents the partial derivatives of the transformation function, providing a way to determine how space distorts under the transformation.

The determinant of the Jacobian, denoted as det\( J_\Phi \), acts as a scaling factor in the integration formula, modifying the size of the small regions as we map them from one domain to another. If there's no scaling effect, this determinant equals one, leading to simplified calculations.

• In our problem, the process was simplified with a Jacobian determinant of 1, meaning that the transformation did not cause any scaling of the area.
• The matrix was easily derived as identity since the transformation function was straightforward: \( \Phi(s, t) = (s, t) \).
• Evaluating the determinant of the Jacobian correctly is crucial since any oversight can drastically alter the outcome of the integral.

Understanding the Jacobian matrix assists in tackling the computational aspects of multivariable calculus effectively, especially as it ensures the calculated integrals remain accurate following transformations.