How do I solve this definite integral?

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In summary, the conversation is about trying to derive Saha's equation, but facing a difficult definite integral. Mathematica gives the solution as 3/2 Zeta[3] = 1.80309, where Zeta is the Riemann zeta function. It is shown that the integral is doable and can be solved using integration by parts and Cauchy's theorem. A more elementary solution is also provided, showing that the integral is equal to 3/2 times the Riemann zeta function evaluated at 3.
  • #1
Ichimaru
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I'm trying to derive Saha's equation, and I've come up against this definite integral, which I can't seem to find anywhere and may not even be doable; I'm not sure.

\begin{equation}

\int\limits_0^\infty \frac{x^{2}}{e^{x}+1}dx

\end{equation}

Can anyone help? Thanks!
 
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  • #3
It's doable. That's essentially the Fermi-Dirac integral F2(0).
 
  • #4
Presumably, traipsing along a different curve in the complex plane, using Cauchy's theorem, might give you the answer.
 
  • #5
Here is a more elementary solution:

Rewrite the given integral as:
$$I=\int_0^{\infty} \frac{x^2e^{-x}}{1+e^{-x}}\,dx$$
Use the substitution ##e^{-x}=t## to obtain:
$$I=\int_0^1 \frac{\ln^2 t}{1+t}\,dt$$
Next, use integration by parts in the following way:
$$I=\left(\ln^2t \ln(1+t)\right|_0^1-2\int_0^1 \frac{2\ln t\ln(1+t)}{t}\,dt$$
Notice that the first term is zero, so we are left with:
$$I=-2\int_0^1 \frac{\ln t \ln(1+t)}{t}\,dt$$
Since,
$$\ln(1+t)=-\sum_{k=1}^{\infty} \frac{(-1)^kt^k}{k}$$
Hence,
$$I=2\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\int_0^1 \frac{t^k\ln t}{t}\,dt$$
It is not difficult to show that
$$\int_0^1 \frac{t^k\ln t}{t}\,dt =-\frac{1}{k^2}$$
From above, we get:
$$I=-2\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$$
It can be easily shown that
$$\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}=-\frac{3}{4}\zeta(3)$$
Hence, the final answer is:
$$I=\frac{3}{2}\zeta(3)$$
which agrees with post #2.

I hope that helps.
 
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  • #6
Cool, Pranav-Arora!
:smile:
 
  • #7
arildno said:
Cool, Pranav-Arora!
:smile:

Thank you! :)
 

1. What is integration and why is it important?

Integration is the process of finding the area under a curve or the accumulation of a quantity over a certain interval. It is important because it allows us to solve a wide range of problems in physics, engineering, economics, and other fields.

2. How do I know which integration method to use?

The method of integration depends on the complexity of the function and the techniques you are familiar with. Some common methods include substitution, integration by parts, and trigonometric substitution. It is best to consult with a math or science expert for guidance on which method to use.

3. What is the difference between definite and indefinite integration?

Definite integration involves finding the numerical value of the integral over a specific interval, while indefinite integration involves finding the general antiderivative of a function. In other words, definite integration gives a number as the answer, while indefinite integration gives a function.

4. Can I use a calculator to integrate?

Yes, most scientific calculators have a built-in integration function. However, it is important to understand the steps and concepts behind integration, rather than relying solely on a calculator.

5. What are some real-world applications of integration?

Integration has many real-world applications, including calculating the area under a graph to find displacement or velocity in physics, finding the total revenue or cost in economics, and determining the probability of an event in statistics. It is also used in engineering for calculating work, power, and other quantities.

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