\int_{-\infty}^\infty {1 \over (4+x^4)} ,dx
To solve the integral ∫(1 / (4 + x^4)) dx from -∞ to ∞, we can use the method of residues. This involves applying the techniques of complex analysis.
Step 1: Determine the poles of the integrand.
Poles are the values of x that make the denominator equal to zero. In this case, solving 4 + x^4 = 0 will give us the values of x where the function is undefined. Factoring the expression, we have (2 + x^2)(2 - x^2) = 0. This simplifies to (x + i)(x - i)(x + 2^(1/4))(x - 2^(1/4)) = 0. Thus, the poles are x = ±i and x = ±2^(1/4).
Step 2: Determine the residues at the poles.
Residues are the coefficients of the Laurent series expansion of the function about each pole. However, for the given integral, the function is even, so we only need to consider the positive poles x = i and x = 2^(1/4).
At x = i, we have the pole of first order. To compute the residue, we evaluate the limit of the function multiplied by (x - i) as x approaches i:
Res(i) = lim(x→i) {(x - i) / (4 + x^4)}
A trick to find the residue is to use L'Hôpital's rule. We differentiate the numerator and the denominator with respect to x:
Res(i) = lim(x→i) {1 / (4x^3)}
= 1 / (4i^3)
= 1 / (-4i)
= -i/4
At x = 2^(1/4), we have another pole of first order. Again, we evaluate the limit as x approaches 2^(1/4):
Res(2^(1/4)) = lim(x→2^(1/4)) {(x - 2^(1/4)) / (4 + x^4)}
Using L'Hôpital's rule, we differentiate the numerator and the denominator:
Res(2^(1/4)) = lim(x→2^(1/4)) {1 / (4x^3)}
= 1 / (4(2^(1/4))^3)
= 1 / (4 × 2^(3/4))
= 1 / (4 × (2^(1/2))^3)
= 1 / (4 × 2 × 2^(1/2))
= 1 / (8 × 2^(1/2))
= 1 / (8 × √2)
= √2 / 64
Step 3: Apply the residue theorem.
The residue theorem states that the integral of a function around a closed curve is equal to 2πi times the sum of the residues inside the curve.
Since our curve encloses only the poles x = i and x = 2^(1/4), the integral can be computed as:
∫(1 / (4 + x^4)) dx = 2πi (Res(i) + Res(2^(1/4)))
∫(1 / (4 + x^4)) dx = 2πi (-i/4 + √2 / 64)
Simplifying the expression, we get:
∫(1 / (4 + x^4)) dx = 1/16π + √2 / (32π)