Applications for Evaluating Real Integrals

Applications for evaluating real integrals using the residue theorem are described in-depth here. In this article, we will look at three different types of integrals and how the residue theorem can be used to evaluate the real integral with the solved examples.

Applications for Evaluating Real Integrals Using Residue Theorem

Case 1

The integrals of the type ∫₀^2π F(cosθ,sinθ) dθ, where F(cosθ, sinθ) is a rational function of cosθ and sinθ.

The substitution method can be used to reduce these integrals to complex line integrals.

z = e^iθ

Hence,

dz = ie^iθdθ,

That is, dθ =dz/iz

In addition,

Cos θ = (½)[z + (1/z)]

Sin θ = (1/2i)[z – (1/z)]

z goes once around the unit circle in the anti-clockwise direction as the θ value varies from 0 to 2π.

where |z| = 1 and C is the unit circle.

The residue theorem can be used to evaluate the integral on the right.

Case 2

The integrals of this type, i.e., ∫_-∞^∞ [f (x)/ F(x)] dx, where f (x) and F (x) are polynomials in x such that [xf(x) /F(x)] → 0 as x → ∞ and F (x) contains no zeros on the Real axis.

Steps:

• Imagine a contour consisting of a huge semi-circle in the upper half of the plane with its center at the origin and its diameter on the real axis.
• Examine whether the numerator’s degree polynomial is bigger than the denominator’s degree polynomial at least by two, resulting in z f(z) → 0 as |z| → 0.
• Locate the poles of f(z) that lie in the plane’s top half.
• Determine the residues located at the poles.
• Therefore, ∫_-∞^∞ f (x) dx = 2πi( sum of residues).

Case 3

The integrals of this type, i.e., ∫_-∞^∞ [f(x)/F(x)] dx, where F (x) has zeros on the real axis.

Let’s have a look at the third case using an example.

Compute ∫₀^∞ [sinx/x] dx

Solution:

Assume that ∫_c (e^iz /z) dz = ∫_c f(z) dz is the contour consisting of

• the real axis from r to R with r being small and R being large.
• the upper half of the huge semi circle C₁, |z| = R
• From -R to -r on the real axis
• the upper half of the smaller semi circle C₂, |z| = r

Cauchy’s Theorem tells us that there are no singularities within the contour C, and we have

Using Jordan’s lemma ∫_c1 f(z) dz = 0

Substituting z = re^iθ and dz = rie^iθdθ, we get ∫_c2 f(z) dz = −πi

Therefore, as r → 0 and R → ∞, we get from equation (1),

On both sides, equating imaginary parts, we get

Solved Example Applications for Evaluating Real Integrals

Example:

Compute

Solution:

Assume a contour consisting of a big semi-circle in the top half of the plane with a center at the origin and a diameter on the real axis.

Therefore,

Z f(z) = [(z³+z²+2z)/(z⁴+10z²+9)] → 0 as |z| → ∞

Furthermore, z⁴ + 10z² + 9 = 0

I.e. (z² + 1) (z² + 9) = 0 results in z = +i, −i, +3i, −3i.

Hence, z = i, 3i are the poles in the upper half.

Therefore, residue at z =i,

Similarly, residue at z = 3i

As a result,

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