Legendre Polynomials Integral Evaluation

The problem requires us to calculate the definite integral of a given polynomial multiplied by the third Legendre polynomial, \(P_3(x)\), over the interval [-1, +1]. We are provided with the Rodrigues formula definition for Legendre polynomials and their orthogonality property.

The integral to be evaluated is: \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \)

P3(x) Calculation from Rodrigues Formula

The Rodrigues formula for Legendre polynomials \(P_n(x)\) is given by:

\(P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2 - 1)^n\)

For \(n=3\), we calculate \(P_3(x)\) by taking the third derivative:

\(P_3(x) = \frac{1}{2^3 3!} \frac{d^3}{dx^3}(x^2 - 1)^3\)

First, expand \((x^2 - 1)^3\):

\((x^2 - 1)^3 = (x^2)^3 - 3(x^2)^2(1) + 3(x^2)(1)^2 - 1^3 = x^6 - 3x^4 + 3x^2 - 1\)

Now, compute the derivatives:

\(\frac{d}{dx}(x^6 - 3x^4 + 3x^2 - 1) = 6x^5 - 12x^3 + 6x\)

\(\frac{d^2}{dx^2}(6x^5 - 12x^3 + 6x) = 30x^4 - 36x^2 + 6\)

\(\frac{d^3}{dx^3}(30x^4 - 36x^2 + 6) = 120x^3 - 72x\)

Substitute this back into the Rodrigues formula for \(P_3(x)\):

\(P_3(x) = \frac{1}{8 \times 6} (120x^3 - 72x)\)

\(P_3(x) = \frac{1}{48} (120x^3 - 72x)\)

Simplifying the expression:

\(P_3(x) = \frac{120}{48}x^3 - \frac{72}{48}x = \frac{5}{2}x^3 - \frac{3}{2}x\)

Thus, the expression for \(P_3(x)\) is \(\frac{1}{2}(5x^3 - 3x)\).

Definite Integral Evaluation Steps

The integral to calculate is \(I = \int_{-1}^{1} (4 + 2x - 3x^2 + 4x^3) P_3(x) dx\). Substitute the expression for \(P_3(x)\) we found:

\(I = \int_{-1}^{1} (4 + 2x - 3x^2 + 4x^3) \frac{1}{2}(5x^3 - 3x) dx\)

We can take the constant factor \(\frac{1}{2}\) outside the integral:

\(I = \frac{1}{2} \int_{-1}^{1} (4 + 2x - 3x^2 + 4x^3) (5x^3 - 3x) dx\)

Next, we expand the product of the two polynomials inside the integral:

\((4 + 2x - 3x^2 + 4x^3) (5x^3 - 3x)\)

• Multiply 4 by \((5x^3 - 3x)\): \(20x^3 - 12x\)
• Multiply \(2x\) by \((5x^3 - 3x)\): \(10x^4 - 6x^2\)
• Multiply \(-3x^2\) by \((5x^3 - 3x)\): \(-15x^5 + 9x^3\)
• Multiply \(4x^3\) by \((5x^3 - 3x)\): \(20x^6 - 12x^4\)

Adding these terms together and combining like powers of x:

\(20x^6 - 15x^5 + (10x^4 - 12x^4) + (20x^3 + 9x^3) - 6x^2 - 12x\)

\(= 20x^6 - 15x^5 - 2x^4 + 29x^3 - 6x^2 - 12x\)

The integral becomes:

\(I = \frac{1}{2} \int_{-1}^{1} (20x^6 - 15x^5 - 2x^4 + 29x^3 - 6x^2 - 12x) dx\)

For integration over the interval [-1, 1], we use the property that for an odd function \(f(-x) = -f(x)\), \(\int_{-1}^1 f(x) dx = 0\), and for an even function \(f(-x) = f(x)\), \(\int_{-1}^1 f(x) dx = 2\int_{0}^1 f(x) dx\). The odd power terms \(x^5, x^3, x^1\) are odd functions, and the even power terms \(x^6, x^4, x^2\) are even functions.

So, the integral simplifies to including only the even power terms:

\(\int_{-1}^{1} (20x^6 - 15x^5 - 2x^4 + 29x^3 - 6x^2 - 12x) dx\)

\(= \int_{-1}^{1} 20x^6 dx + \int_{-1}^{1} (-2x^4) dx + \int_{-1}^{1} (-6x^2) dx + \int_{-1}^{1} (\text{odd terms}) dx\)

\(= 2 \int_{0}^{1} 20x^6 dx + 2 \int_{0}^{1} (-2x^4) dx + 2 \int_{0}^{1} (-6x^2) dx + 0\)

\(= 2 \left[ \frac{20x^{6+1}}{6+1} \right]_0^1 + 2 \left[ \frac{-2x^{4+1}}{4+1} \right]_0^1 + 2 \left[ \frac{-6x^{2+1}}{2+1} \right]_0^1\)

\(= 2 \left[ \frac{20x^7}{7} \right]_0^1 + 2 \left[ \frac{-2x^5}{5} \right]_0^1 + 2 \left[ \frac{-6x^3}{3} \right]_0^1\)

Evaluate the definite integral from 0 to 1:

\(= 2 \left( \frac{20(1)^7}{7} - \frac{20(0)^7}{7} \right) + 2 \left( \frac{-2(1)^5}{5} - \frac{-2(0)^5}{5} \right) + 2 \left( \frac{-6(1)^3}{3} - \frac{-6(0)^3}{3} \right)\)

\(= 2 \left( \frac{20}{7} \right) + 2 \left( \frac{-2}{5} \right) + 2 \left( -2 \right)\)

\(= \frac{40}{7} - \frac{4}{5} - 4\)

To combine these terms, find a common denominator, which is 35:

\(= \frac{40 \times 5}{7 \times 5} - \frac{4 \times 7}{5 \times 7} - \frac{4 \times 35}{1 \times 35}\)

\(= \frac{200}{35} - \frac{28}{35} - \frac{140}{35}\)

\(= \frac{200 - 28 - 140}{35} = \frac{200 - 168}{35} = \frac{32}{35}\)

Remember the factor of \(\frac{1}{2}\) that was taken outside the integral initially:

\(I = \frac{1}{2} \times \frac{32}{35} = \frac{16}{35}\)

Legendre Orthogonality Approach

Alternatively, we can use the orthogonality property of Legendre polynomials. Any polynomial \(Q(x)\) of degree \(N\) can be expanded as a linear combination of Legendre polynomials \(P_n(x)\) up to degree \(N\): \(Q(x) = \sum_{n=0}^N c_n P_n(x)\). In this problem, \(Q(x) = 4 + 2x - 3x^2 + 4x^3\) is a polynomial of degree 3.

The integral is \(\int_{-1}^1 Q(x) P_3(x) dx = \int_{-1}^1 (\sum_{n=0}^3 c_n P_n(x)) P_3(x) dx\). Due to the orthogonality condition \(\int_{-1}^1 P_n(x) P_m(x) dx = \frac{2}{2n+1} \delta_{nm}\), where \(\delta_{nm}\) is the Kronecker delta (\(\delta_{nm}=1\) if \(n=m\) and 0 otherwise), only the term where \(n=3\) survives:

\(\int_{-1}^1 Q(x) P_3(x) dx = c_3 \int_{-1}^1 P_3(x) P_3(x) dx = c_3 \frac{2}{2(3)+1} = c_3 \frac{2}{7}\)

To find the coefficient \(c_3\), we compare the leading terms of \(Q(x)\) and the expansion. \(Q(x) = 4x^3 - 3x^2 + 2x + 4\). The leading term is \(4x^3\).

The leading term of \(P_n(x)\) is given by \(\frac{(2n)!}{2^n (n!)^2} x^n\). For \(n=3\), the leading term of \(P_3(x)\) is \(\frac{(6)!}{2^3 (3!)^2} x^3 = \frac{720}{8 \times 36} x^3 = \frac{720}{288} x^3 = \frac{5}{2} x^3\).

Equating the leading coefficients of \(Q(x)\) and \(c_3 P_3(x)\) (which dominates the expansion for \(x^3\)):

\(4x^3 = c_3 \left(\frac{5}{2}x^3\right) + \text{lower order terms}\)

\(4 = c_3 \frac{5}{2}\)

Solving for \(c_3\):

\(c_3 = 4 \times \frac{2}{5} = \frac{8}{5}\)

Now substitute this value of \(c_3\) into the integral expression \(c_3 \frac{2}{7}\):

\(I = \frac{8}{5} \times \frac{2}{7} = \frac{16}{35}\)

Both methods yield the same result, confirming the calculation.

Integral Final Result

The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is \(\frac{16}{35}\).