Evaluate the triple integral over all of R³ of exp(−alpha times (x² + y² + z²)) multiplied by the sum: a cos(xξ + yη + zζ) plus b sin(xξ + yη + zζ) plus c times (x² + y² + z²).

Step-by-step solution

1. 1

   Move to vector notation and split into three integrals. Write $\mathbf{x} = (x,y,z)$, $\mathbf{k} = (\xi,\eta,\zeta)$, $r = \|\mathbf{x}\|$ and $k = \|\mathbf{k}\|$. Then $x\xi + y\eta + z\zeta = \mathbf{x}\cdot\mathbf{k}$ and

   $I = aI_1 + bI_2 + cI_3$

   with $I_1 = \iiint e^{-\alpha r^2}\cos(\mathbf{x}\cdot\mathbf{k})\,d^3x$, $I_2$ the same with sine, and $I_3 = \iiint e^{-\alpha r^2} r^2\,d^3x$. The condition $\alpha > 0$ is what makes all three converge.

2. 2

   Kill the sine term by parity. Under $\mathbf{x} \mapsto -\mathbf{x}$ the Gaussian $e^{-\alpha r^2}$ is unchanged while $\sin(\mathbf{x}\cdot\mathbf{k})$ flips sign, so the integrand is odd over a domain symmetric about the origin:

   $I_2 = 0$

   The coefficient $b$ therefore never appears in the answer.

3. 3

   Evaluate the cosine term as a Fourier transform. The standard 3-D Gaussian transform is

   $\iiint_{\mathbb{R}^3} e^{-\alpha r^2} e^{i\mathbf{x}\cdot\mathbf{k}}\,d^3x = \left(\frac{\pi}{\alpha}\right)^{3/2} e^{-k^2/(4\alpha)}$

   (each Cartesian direction contributes $\sqrt{\pi/\alpha}\,e^{-k_i^2/4\alpha}$ after completing the square). Taking real parts — the imaginary part is exactly $I_2 = 0$ — gives

   $I_1 = \left(\frac{\pi}{\alpha}\right)^{3/2} e^{-k^2/(4\alpha)}$

4. 4

   Evaluate the r² term in spherical coordinates. The integrand depends only on $r$, so $d^3x = 4\pi r^2\,dr$:

   $I_3 = 4\pi\int_0^{\infty} r^4 e^{-\alpha r^2}\,dr$

   The substitution $u = \alpha r^2$ turns this into a gamma integral:

   $\int_0^\infty r^4 e^{-\alpha r^2}dr = \frac{\Gamma(5/2)}{2\alpha^{5/2}} = \frac{3\sqrt{\pi}}{8\alpha^{5/2}}$

   using $\Gamma(5/2) = \tfrac{3}{4}\sqrt{\pi}$. Hence $I_3 = \dfrac{3\pi^{3/2}}{2\alpha^{5/2}}$.

5. 5

   Assemble the result.

   $I = a\left(\frac{\pi}{\alpha}\right)^{3/2} e^{-(\xi^2+\eta^2+\zeta^2)/(4\alpha)} + \frac{3c\,\pi^{3/2}}{2\alpha^{5/2}}$

   Note the structure: the $a$-part decays in $k$ (a Gaussian in frequency space), while the $c$-part is a pure constant independent of $\mathbf{k}$.

6. 6

   Verify numerically. With $\alpha = 0.7$, $\mathbf{k} = (0.3, -0.5, 1.1)$, $a = 1.3$, $b = 2.1$, $c = 0.9$, the closed form gives $25.4420947$. A $90$-point Gauss-Legendre grid on $[-6,6]^3$ (the Gaussian is negligible beyond that) gives $25.4420947$ — agreement to eight digits, including the fact that the $b = 2.1$ term contributes nothing.