Question

Express each of the following integrals as a $\mathrm{\Gamma }$ function. By computer, evaluate numerically both the $\mathrm{\Gamma }$ function and the original integral. ${\int }_0^{\mathrm{\infty }}{x}^{-1/3}{e}^{-8x}dx$

Short answer

The integral ${\int }_0^{\text{∞}}{x}^{-1/3}{e}^{-8x}dx$ is $2^{-2}\mathrm{\Gamma }(2/3)\approx 0.3385$.

Step-by-step solution

Identify the integral form

The given integral is ${\int }_0^{\text{∞}}{x}^{-1/3}{e}^{-8x}dx$. We need to express it in terms of the Gamma function $\mathrm{\Gamma }(t)$.

Match to Gamma Function Form

Recall the Gamma function form: $Gamma(t)={\int }_0^{\text{∞}}{x}^{t-1}{e}^{-x}dx$ . Observe the exponents and the exponential factor in the given integral.

Make a Substitution

Let's make the substitution $u=8x$. Therefore, $dx=\frac{du}{8}$. The integral becomes ${\int }_0^{\text{∞}}(u/8{)}^{-1/3}{e}^{-u}(1/8)du$.

Simplify the Integral

Simplify the integral expression ${\int }_0^{\text{∞}}{8}^{1/3}{u}^{-1/3}{e}^{-u}\frac{1}{8}du$. Using $8^{1/3}=2$ and $1/8=2^{-3}$, we get $2^{1-3}{\int }_0^{\text{∞}}{u}^{-1/3}{e}^{-u}du$.

Express in Terms of Gamma Function

Recognize that ${\int }_0^{\text{∞}}{u}^{-1/3}{e}^{-u}du$ represents $\mathrm{\Gamma }(-1/3+1)=\mathrm{\Gamma }(2/3)$. Thus, the original integral becomes $2^{-2}\mathrm{\Gamma }(2/3)$.

Compute the Value

Using numerical methods or computer software, evaluate $\mathrm{\Gamma }(2/3)\approx 1.354$. So, the integral evaluates to $2^{-2}\mathrm{\Gamma }(2/3)\approx 1.354/4=0.3385$.

Key concepts

Integral Substitution

Integral substitution is a technique used in calculus to simplify the process of evaluating integrals.
It involves changing the variable of integration, leading to a simpler form of the integral that is easier to evaluate.
In our original problem, we made the substitution $u=8x$, which transformed the integral into a form that resembles the Gamma function.
When making a substitution, always remember to adjust the differential element accordingly. In this case, $du=8dx$ leads to $dx=\frac{du}{8}$.
This step is crucial because it ensures the new integral is correctly set up. After substitution, if the integral resembles a known form (like the Gamma function), it can be evaluated more easily.

Definite Integrals

Definite integrals represent the area under a curve between two specific limits.
In our example, the integral ${\int }_0^{\mathrm{\infty }}{x}^{-1/3}{e}^{-8x}dx$ has the limits from 0 to infinity.
Evaluating definite integrals often involves techniques like substitution or recognizing standard forms, as we did with the Gamma function.
Unlike indefinite integrals, definite integrals yield a numerical value, showing the total 'accumulation' between the limits.
Understanding how to manipulate and transform these integrals is key to solving many problems in calculus.

Numerical Integration

Sometimes, integrals can't be solved analytically, and numerical methods are used instead.
Methods like the Trapezoidal Rule, Simpson's Rule, or more advanced techniques (e.g., Gaussian quadrature) are common.
In our example, we used numerical software to evaluate $\mathrm{\Gamma }(2/3)$, approximately equal to 1.354.
Numerical integration provides an approximate solution, which is often necessary for complex or non-standard functions.
It is essential to choose an appropriate method based on the function's properties and the desired accuracy.