Solving Improper Integrals Using Exponential and Cosine Functions

Improper integrals are a crucial part of mathematical analysis. This article explores the evaluation of two specific improper integrals, namely (int_{0}^{infty} frac{sin x}{x^2} dx) and (int_{0}^{infty} frac{cos x}{x^2} dx). We will present these solutions in terms of the exponential integral function and Laplace transforms, providing a detailed derivation process.

Introduction to Improper Integrals

Improper integrals are integrals that are unbounded either in the limits of integration or have non-integrable divergences. These integrals can be evaluated using various techniques from advanced calculus and special functions.

Evaluation of (int_{0}^{infty} frac{sin x}{x^2} dx)

The integral (int_{0}^{infty} frac{sin x}{x^2} dx) can be solved using the exponential integral function (E_i(x)).

The integral can be rewritten using the exponential integral function:

[int_{0}^{infty} frac{sin x}{x^2} dx frac{E_1(1) - e^{-2} E_{-1}(1)}{2e}]

Here, (E_1(x)) and (E_{-1}(x)) are the first and second forms of the exponential integral function, respectively. Evaluating this expression yields:

[ frac{1.895 - 7.389 cdot (-0.2194)}{2 cdot 2.718} approx 0.647]

Evaluation of (int_{0}^{infty} frac{cos x}{x^2} dx)

For the integral (int_{0}^{infty} frac{cos x}{x^2} dx), a detailed approach has been provided by Laplace in 1811. Inspired by his work, we aim to derive the solution using Laplace transforms and similar techniques.

Proof Using Laplace Transforms

The integral (int_{0}^{infty} 2z e^{-x^2 z^2} dz) is first evaluated:

[int_{0}^{infty} 2z e^{-x^2 z^2} dz -left[frac{e^{-x^2 z^2}}{x^2}right]_0^{infty} frac{1}{x^2}]

Using this result, we can calculate the desired integral:

[int_{0}^{infty} frac{cos x}{x^2} dx int_{0}^{infty} int_{0}^{infty} cos x cdot 2z e^{-x^2 z^2} dxdz]

Now, we simplify the inner integral using the known result:

[ int_{0}^{infty} 2z e^{-z^2} left[int_{0}^{infty} cos x cdot e^{-x^2 z^2} dxright] dz]

From Appendix A, we know that:

[int_{0}^{infty} cos x cdot e^{-x^2 z^2} dx frac{sqrt{pi}}{2z} e^{-frac{1}{4z^2}}]

Substituting this into the integral, we get:

[ sqrt{pi} int_{0}^{infty} e^{-left(z^2 - frac{1}{4z^2}right)} dz]

Using another substitution, we obtain:

[ sqrt{pi} frac{sqrt{pi}}{2} e^{-1} frac{pi}{2e} 0.578]

Thus, the value of the integral is (0.578).

Appendix A: Detailed Proof for (int_{0}^{infty} cos ax cdot e^{-x^2 z^2} dx)

We start by defining the integral:

[I_a int_{0}^{infty} cos ax cdot e^{-x^2 z^2} dx]

Using the technique of differentiation under the integral sign, we differentiate with respect to (a):

[frac{dI}{da} -int_{0}^{infty} x sin ax cdot e^{-x^2 z^2} dx]

After integration by parts, we obtain the differential equation:

[frac{dI}{da} -frac{a}{2z^2} I_a]

Solving this differential equation, we get:

[I_a A e^{-frac{a^2}{4z^2}}]

Using the boundary condition (I_a int_{0}^{infty} e^{-x^2 z^2} dx frac{sqrt{pi}}{2}), we find (A frac{sqrt{pi}}{2}). Thus, the solution is:

[I_a frac{sqrt{pi}}{2} e^{-frac{a^2}{4z^2}}]

As a result, when (a 1), we get:

[int_{0}^{infty} cos x cdot e^{-x^2 z^2} dx frac{sqrt{pi}}{2} e^{-frac{1}{4z^2}}]

Appendix B: Detailed Proof for (int_{0}^{infty} e^{-frac{a^2}{x^2}} dx)

We consider the integral:

[I_a int_{0}^{infty} e^{-frac{a^2}{x^2}} dx]

By differentiating with respect to (a), we obtain:

[frac{dI}{da} -2a int_{0}^{infty} frac{1}{x^2} e^{-frac{a^2}{x^2}} dx]

Using the substitution (x frac{a}{y}) and simplifying, we get:

[frac{dI}{da} -2a int_{0}^{infty} e^{-frac{a^2}{x^2}} dx -2I_a]

Solving this differential equation, we find:

[I_a A e^{-2a}]

Using the boundary condition (I_a int_{0}^{infty} e^{-x^2} dx frac{sqrt{pi}}{2}), we determine (A frac{sqrt{pi}}{2}). Therefore, the solution is:

[I_a frac{sqrt{pi}}{2} e^{-2a}]

Finally, we evaluate the integral:

[int_{0}^{infty} e^{-frac{1}{4z^2}} dz frac{1}{2} frac{sqrt{pi}}{2} e^{-1}]

This completes the detailed evaluation of the given improper integrals using the exponential integral function and Laplace transforms.