- #1
Euge
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Evaluate the definite integral $$\int_0^\infty \frac{x^2 + 1}{x^4 + 1}\, dx$$
Using ##(2n)!! = 2n (2n-2) \cdots 2 = 2^n n!##,anuttarasammyak said:Another approach:
transforming x by
[tex]x=\tan\theta[/tex]
The integral is
[tex]\int_0^\pi \frac{d\phi}{1+\cos^2 \phi}=\sum_{n=0}^\infty (-1)^n \int_0^\pi \cos^{2n}\phi d\phi=\pi \sum_{n=0}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!}[/tex]
where
[tex]\phi=2\theta[/tex]
The series converges slowly. The sum from n=0 upto 50,000 gives the estimation between 0.7058
and 0.7083 for the expected ##1/\sqrt{2}##=0.70710...
julian said:Method #1
Using ##u = x^4##,
\begin{align*}
\int_0^\infty \dfrac{x^2+1}{x^4+1} dx = \frac{1}{4} \int_0^\infty \dfrac{u^{-1/4}+u^{-3/4}}{u+1} du \qquad (*)
\end{align*}
Here:
www.physicsforums.com/threads/exponential-type-integrals.1046843/
in post #12 I proved that, if ##0< a < 1##, then
$$
\int_0^\infty \frac{u^{a-1}}{1+u} du = \frac{\pi}{\sin a \pi}
$$
Using this in ##(*)##, we have
\begin{align*}
\int_0^\infty \dfrac{x^2+1}{x^4+1} dx & = \frac{\pi}{4} \left( \frac{1}{\sin 3 \pi / 4} + \frac{1}{\sin \pi / 4} \right)
\nonumber \\
& = \frac{\pi}{\sqrt{2}}
\end{align*}
Yes, you get out the additional result:pasmith said:In method #1, doing the contour integration directly also gets the value of [tex]
\int_0^\infty \frac{1 - x^2}{1 + x^4}\,dx.[/tex]
A definite integral of a rational function is a mathematical concept that represents the area under the curve of the function between two given points on the x-axis. It is denoted by ∫ab f(x) dx, where a and b are the lower and upper limits of integration, and f(x) is the rational function.
The definite integral of a rational function can be calculated using the fundamental theorem of calculus, which states that the integral of a function can be found by evaluating its antiderivative at the upper and lower limits of integration and taking the difference between the two values. In the case of a rational function, the antiderivative can be found by using integration techniques such as substitution or partial fractions.
The definite integral of a rational function has several properties, including linearity, meaning that the integral of a sum of rational functions is equal to the sum of their individual integrals. It also follows the property of the definite integral of a constant, which states that the integral of a constant multiplied by a function is equal to the constant multiplied by the integral of the function. Additionally, the definite integral of a rational function is affected by the choice of limits of integration, with different limits resulting in different values for the integral.
The definite integral of a rational function has many real-world applications, including calculating the area under a curve in physics and engineering problems. It is also used in economics to calculate the total profit or loss of a business over a certain period of time. In statistics, the definite integral of a rational function is used to find the probability of a continuous random variable falling within a certain range of values.
Yes, there are some limitations and restrictions when calculating the definite integral of a rational function. For example, if the function has a vertical asymptote within the limits of integration, the integral will not exist. Similarly, if the function has a discontinuity within the limits of integration, the integral will also not exist. Additionally, some rational functions may be too complex to integrate using traditional techniques, requiring the use of numerical methods such as the trapezoidal rule or Simpson's rule.