Question 705033

We are given the integral:

$I = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp[-(x^2 + y^2 + 2axy)] dx dy$

We need to eliminate the coupled term $(2axy)$ from the exponent. For this, we use a standard linear transformation that rotates the axes by $45^\circ$:

Step 1: Define the Transformation Let the new variables be $u$ and $v$, where:

$x = \frac{u + v}{\sqrt{2}}$

$y = \frac{u - v}{\sqrt{2}}$

Step 2: Calculate the Jacobian Now, we differentiate $x$ and $y$ with respect to $u$ and $v$ to form the Jacobian matrix:

$J = \det \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}$

Solving the determinant:

$J = \left(\frac{1}{\sqrt{2}} \times -\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\right) = -\frac{1}{2} - \frac{1}{2} = -1$

The area scale factor is always positive, so we take its absolute value:

$|J| = |-1| = 1$

This means in the new coordinate system, $dx dy = du dv$. (Since this is a pure rotation, the area neither stretches nor shrinks).

Step 3: Convert the Exponent to the New Variables Now, substitute the transformation formulas into the original exponent $x^2 + y^2 + 2axy$:

First, let's evaluate $x^2 + y^2$:

$x^2 + y^2 = \left(\frac{u + v}{\sqrt{2}}\right)^2 + \left(\frac{u - v}{\sqrt{2}}\right)^2 = \frac{u^2 + v^2 + 2uv + u^2 + v^2 - 2uv}{2} = u^2 + v^2$

Next, evaluate $2xy$:

$2xy = 2 \left(\frac{u + v}{\sqrt{2}}\right) \left(\frac{u - v}{\sqrt{2}}\right) = 2 \left(\frac{u^2 - v^2}{2}\right) = u^2 - v^2$

Substituting these back into the exponent:

$(u^2 + v^2) + a(u^2 - v^2)$

$u^2(1 + a) + v^2(1 - a)$

Notice that the cross-term (coupled term) has completely vanished!

Step 4: Evaluate the Integral and Check Convergence Conditions The limits of integration remain from $-\infty$ to $\infty$. The new integral becomes:

$I = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp[-u^2(1 + a) - v^2(1 - a)] du dv$

Since $u$ and $v$ are now independent, we can split the integral into two separate standard 1D Gaussian integrals:

$I = \left( \int_{-\infty}^{\infty} \exp[-u^2(1 + a)] du \right) \times \left( \int_{-\infty}^{\infty} \exp[-v^2(1 - a)] dv \right)$

A standard Gaussian integral $\int \exp(-\alpha z^2) dz$ is finite (convergent) only if the constant $\alpha > 0$.

Therefore, we obtain two conditions for the two integrals:

1. For the first integral: $1 + a > 0 \implies a > -1$
2. For the second integral: $1 - a > 0 \implies a < 1$

Both conditions are simultaneously satisfied only when the value of $a$ lies between $-1$ and $1$.

Final Answer: The full range of values of $a$ for which the integral is finite is $-1 < a < 1$.