# Definite integral involving a lot of exponentials.

• MHB
• caffeinemachine
In summary, the given problem involves evaluating the integral $\int_{0}^\infty \frac{e^{3x} - e^x}{x(e^x + 1)(e^{3x} + 1)}\ dx$. By substituting $y=e^x$ and changing the bounds to $\int_0^1$, the integral can be rewritten as $\int_0^1 \frac{y-1}{(y^3+1)\log{y}}\,\mathrm{d}y$. By expanding the geometric series and switching the order of integration and series, we can transform the integral into an infinite series. This can then be calculated by turning it into a product using the cosine product
caffeinemachine
Gold Member
MHB
Problem: Evaluate
$$\int_{0}^\infty \frac{e^{3x} - e^x}{x(e^x + 1)(e^{3x} + 1)}\ dx$$

Attempt.

I substituted $y=e^x$, thus $dx = dy/y$, which turns the above integral to

$$\int_{1}^\infty \frac{y^2 - 1}{(\log y)(y+1)(y^3+1)}\ dy = \int_{1}^\infty \frac{y-1}{(\log y) (y^3+1)} \ dy$$

I am unable to make progress.

Thanks.

1. Sub $y \mapsto \frac{1}{y}$ to change the bounds.

2. Note that $\displaystyle \frac{y-1}{\log{y}}= \int_0^1 y^t \, \mathrm{d}t$.

3. Switch the order of the double integral

4. Expand the geometric series

5. Switch the order of integral and series

6. You're left with an infinite series.

This series can be calculated by turning it into a product by putting it inside the log and using

$\displaystyle \cos\pi z=\prod_{n \ge 0 }\left(1-\frac{4z^2}{(2n+1)^2}\right)$

I'm too tired to write out detailed answer now, but that should be the gist of it.

Letting $y \mapsto \frac{1}{y}$ we get $\displaystyle I = \int_0^1 \frac{y-1}{(y^3+1)\log{y}}\,\mathrm{d}y$; since $\displaystyle \frac{y-1}{\log{y}}= \int_0^1 y^t \, \mathrm{d}t$, we have

\begin{aligned} I & = \int_0^1 \int_0^1 \frac{y^t}{y^3+1}\,\mathrm{d}t\,\mathrm{d}y = \int_0^1 \int_0^1 \frac{y^t}{y^3+1}\,\mathrm{d}y\,\mathrm{d}t\end{aligned}

Since $\displaystyle \frac{1}{1+y^3} = \sum_{k \ge 0} y^{3k}$ uniformly over $y \in (0, 1)$, we have

\begin{aligned} I & = \int_0^1 \int_0^1 \sum_{k \ge 0}(-1)^k y^{3k+t}\,\mathrm{d}y\,\mathrm{d}t \\& = \sum_{k \ge 0}(-1)^k\int_0^1 \int_0^1 y^{3k+t}\,\mathrm{d}y\,\mathrm{d}t \\& = \sum_{k \ge 0} (-1)^k \log\left(\frac{3k+2}{3k+1}\right) \\& =\log\left(\prod_{k \ge 0} \bigg[\frac{3k+2}{3k+1}\bigg]^{(-1)^k}\right) \\&\end{aligned}

To calculate the product we have we have:

\begin{aligned}\displaystyle \prod_{k \ge 0} \bigg[\frac{3k+2}{3k+1}\bigg]^{(-1)^k} & = \prod_{k \ge 0} \bigg[\frac{6k+2}{6k+1}\bigg]\bigg[\frac{6k+4}{6k+5}\bigg] \\& = \prod_{k \ge 0} \frac{(6k+2)(6k+4)}{(6k+1)(6k+5)} \\& = \frac{\displaystyle \prod \left[1-4/6^2/(2k+1)^2\right]}{\displaystyle \prod \left[1-4/3^2/(2k+1)^2\right]} \\& = \frac{\cos{\frac{1}{6}\pi}}{\cos{\frac{1}{3} \pi}} \\& = \sqrt{3}.\end{aligned}Hence $I = \log{\sqrt{3}}.$

Last edited:

## 1. What is a definite integral?

A definite integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value of a function over a specific interval.

## 2. What are exponentials?

Exponentials are mathematical expressions in the form of ax, where a is a constant and x is a variable. They are often used to model growth or decay in various natural and scientific phenomena.

## 3. How do you solve a definite integral involving exponentials?

To solve a definite integral involving exponentials, you can use techniques such as substitution, integration by parts, or partial fractions. It is important to follow the fundamental rules of integration and pay attention to the properties of exponentials.

## 4. What are some real-life applications of definite integrals involving exponentials?

Definite integrals involving exponentials are commonly used in physics, chemistry, and engineering to model and analyze various processes such as population growth, radioactive decay, and chemical reactions. They are also used in economics and finance to calculate compound interest and investment growth.

## 5. Can definite integrals involving exponentials have negative values?

Yes, definite integrals involving exponentials can have negative values. This can occur when the function being integrated has negative values over the given interval, or when the area under the curve is below the x-axis. However, the definite integral itself represents the total signed area, so it can also be positive or zero.

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