Solve differential equations

Oct 13, 2025 by leolavanderia.com.br 122 views

Introduction

In the realm of calculus, definite integrals often present a challenge to evaluate, especially when they involve complex functions. One such example is the integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ . In this article, we will delve into the process of finding a closed form for this integral, which ultimately leads to an expression involving hypergeometric functions.

Background and Motivation

The given integral was initially part of a harmonic sum, which was reduced to an expression involving two hypergeometric functions, $\mathcal{H}_1$ and $\mathcal{H}_2$ . The first hypergeometric function, $\mathcal{H}_1$ , is defined as:

\mathcal{H}_1(a, b; c; z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}

where $(a)_n$ denotes the Pochhammer symbol. The second hypergeometric function, $\mathcal{H}_2$ , is defined as:

\mathcal{H}_2(a, b; c; z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}

Our goal is to find a closed form for the integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ , which will ultimately lead to an expression involving these hypergeometric functions.

Methodology

To tackle this problem, we will employ a combination of mathematical techniques, including substitution, integration by parts, and the use of hypergeometric functions. We will start by making a substitution to simplify the integral, followed by integration by parts to reduce the order of the integral.

Substitution

Let's make the substitution $x^4 = 2t$ . This will simplify the integral and make it more manageable.

\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x = \int_0^1 \frac{x^2 \, \ln x }{(1 - t)^{5/4}} \mathrm{d}x

Now, we can rewrite the integral in terms of $t$ :

\int_0^1 \frac{x^2 \, \ln x }{(1 - t)^{5/4}} \mathrm{d}x = \int_0^1 \frac{(2t)^{1/2} \, \ln (2t)^{1/4}}{(1 - t)^{5/4}} \mathrm{d}t

Integration by Parts

Next, we will use integration by parts to reduce the order of the integral. We will choose $u = \ln (2t)^{1/4}$ and $dv = \frac{(2t)^{1/2}}{(1 - t)^{5/4}} \mathrm{d}t$ .

\int_0^1 \frac{(2t)^{1/2} \, \ln (2t)^{1/4}}{(1 - t)^{5/4}} \mathrm{d}t = \int_0^1 \ln (2t)^{1/4} \, \frac{(2t)^{1/2}}{(1 - t)^{5/4}} \mathrm{d}t

Now, we can integrate by parts:

\int_0^1 \ln (2t)^{1/4} \, \frac{(2t)^{1/2}}{(1 - t)^{5/4}} \mathrm{d}t = \left[ \ln (2t)^{1/4} \, \frac{(2t)^{1/2}}{(1 - t)^{5/4}} \right]_0^1 - \int_0^1 \frac{(2t)^{1/2}}{(1 - t)^{5/4}} \, \mathrm{d}t

Hypergeometric Functions

The remaining integral can be expressed in terms of hypergeometric functions. We will use the definition of the hypergeometric function $\mathcal{H}_1$ to rewrite the integral:

\int_0^1 \frac{(2t)^{1/2}}{(1 - t)^{5/4}} \mathrm{d}t = \mathcal{H}_1\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; \frac{1}{2}\right)

Closed Form

Now, we can combine the results to obtain a closed form for the original integral:

\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x = \mathcal{H}_1\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; \frac{1}{2}\right)

This is the closed form for the given integral, which involves the hypergeometric function $\mathcal{H}_1$ .

Conclusion

In this article, we have derived a closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ . The process involved a combination of mathematical techniques, including substitution, integration by parts, and the use of hypergeometric functions. The final result is an expression involving the hypergeometric function $\mathcal{H}_1$ , which provides a closed form for the original integral.

References

[1] Abramowitz, M., and Stegun, I. A. (1964). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standards Applied Mathematics Series, 55.
[2] Bailey, W. N. (1935). Generalized Hypergeometric Series. Cambridge University Press.
[3] Slater, L. J. (1966). Generalized Hypergeometric Functions. Cambridge University Press.

Introduction

In our previous article, we derived a closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ . This integral involves the hypergeometric function $\mathcal{H}_1$ , which is a fundamental concept in mathematics. In this article, we will answer some of the most frequently asked questions about this topic.

Q: What is the hypergeometric function $\mathcal{H}_1$ ?

A: The hypergeometric function $\mathcal{H}_1$ is a special function that is defined as:

\mathcal{H}_1(a, b; c; z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}

where $(a)_n$ denotes the Pochhammer symbol.

Q: What is the Pochhammer symbol?

A: The Pochhammer symbol is a mathematical notation that is used to represent the product of consecutive integers. It is defined as:

(a)_n = a(a+1)(a+2)\cdots(a+n-1)

Q: How is the hypergeometric function $\mathcal{H}_1$ used in mathematics?

A: The hypergeometric function $\mathcal{H}_1$ is used in a wide range of mathematical applications, including:

Special functions: The hypergeometric function $\mathcal{H}_1$ is a special function that is used to represent a wide range of mathematical functions, including the gamma function, the beta function, and the error function.
Calculus: The hypergeometric function $\mathcal{H}_1$ is used in calculus to represent the solution to a wide range of differential equations, including the heat equation, the wave equation, and the Laplace equation.
Number theory: The hypergeometric function $\mathcal{H}_1$ is used in number theory to represent the solution to a wide range of Diophantine equations, including the Pell equation and the Markov equation.

Q: What is the significance of the closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ ?

A: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ is significant because it provides a precise and accurate solution to a wide range of mathematical problems. This solution can be used to:

Solve differential equations: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ can be used to solve a wide range of differential equations, including the heat equation, the wave equation, and the Laplace equation.
Represent special functions: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ can be used to represent a wide range of special functions, including the gamma function, the beta function, and the error function.
Solve Diophantine equations: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ can be used to solve a wide range of Diophantine equations, including the Pell equation and the Markov equation.

Q: How can I use the closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ in my own research?

A: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ can be used in a wide range of mathematical applications, including:

Calculus: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ can be used to solve a wide range of differential equations, including the heat equation, the wave equation, and the Laplace equation.
Number theory: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ can be used to solve a wide range of Diophantine equations, including the Pell equation and the Markov equation.
Special functions: The closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ can be used to represent a wide range of special functions, including the gamma function, the beta function, and the error function.

Conclusion

In this article, we have answered some of the most frequently asked questions about the closed form for the definite integral $\int_0^1 \frac{x^2 \, \ln x }{\left(1 - \frac{x^4}{2} \right)^{5/4}} \mathrm{d}x$ . This solution is significant because it provides a precise and accurate solution to a wide range of mathematical problems. We hope that this article has been helpful in providing a better understanding of this topic.