Show that \(\int_{-1}^{1} P_{l}(x) d x=0, l>0 .\) Hint: Consider \(\int_{-1}^{1} P_{l}(x) P_{0}(x) d x.\)

Orthogonality is a concept that helps simplify complex problems by reducing interference between functions. Think of it like vectors: just as perpendicular vectors don’t affect each other’s directions, orthogonal functions don’t interfere with each other’s integrals.

For Legendre polynomials, the orthogonality property means that the integral of the product of different polynomials (over the interval \([-1, 1]\)) is zero. Mathematically, this is written as:
\[ \int_{-1}^{1} P_{l}(x) P_{m}(x) \ dx = 0 \ if \ l \eq \ m \]
Orthogonality simplifies many calculations, particularly in physics and engineering, where Legendre polynomials often show up in solutions to differential equations.

Integrals help in finding the area under a curve, but they also hold significance in understanding properties like orthogonality. Here, the key property we use is the orthogonality of Legendre polynomials over the interval \([-1, 1]\).

Due to this property:
\[ \int_{-1}^{1} P_{l}(x) P_{0}(x) \ dx = 0 \ for \ l > 0 \]
Knowing \[ P_{0}(x) = 1 \] simplifies our problem to: \[ \int_{-1}^{1} P_{l}(x) \ dx = 0 \ for \ l > 0 \]
This shows that the integral of any Legendre polynomial of order greater than zero is zero over the interval \([-1, 1]\), leveraging the orthogonality property.

The zeroth Legendre polynomial, \ P_{0}(x)\, is the simplest of the Legendre polynomials. It is defined as: \[ P_{0}(x) = 1 \]
This is a constant function. The simplicity of \[P_{0}(x)\] helps in utilizing many integral properties.

When dealing with the orthogonality of Legendre polynomials, \[P_{0}(x)\] plays a pivotal role because:
\[\int_{-1}^{1} P_{l}(x) P_{0}(x) \ dx = 0 \ for \ l > 0 \]
Simplifies to: \[\int_{-1}^{1} P_{l}(x) \ dx = 0 \ for \ l > 0 \]
Understanding \ P_{0}(x)\ makes it easier to grasp the behavior and properties of higher-order Legendre polynomials.