Finding Multiple Ways to Integrate

[Comeback City]

Hello everyone!

Today in my BC Calculus (calc 2) class, we spent time trying to find ways to integrate the same function (all methods we're fluent with, but are reviewing for the upcoming AP test)...
(I apologize in advance that I do not know how to use LaTex, but am trying to learn)

f(x) = 1/[(e^x)+1]

Overall, we were able to integrate using simple methods such as trigonometric substitution, u-substitution, partial fraction decomposition, and with another algebraic method.

Being me, I didn't think this was enough, and attempted to solve this using a series. Below is my work...

My question... Is there a way to continue this series and prove it is equivalent to the algebraic form found using the other integration methods? That is, besides just saying they come from the same integral.

Thank you in advance

 

[member 587159]

Did you justify the change of the integral and the series?

 

[pasmith]

Hello everyone!

Today in my BC Calculus (calc 2) class, we spent time trying to find ways to integrate the same function (all methods we're fluent with, but are reviewing for the upcoming AP test)...
(I apologize in advance that I do not know how to use LaTex, but am trying to learn)

f(x) = 1/[(e^x)+1]

Overall, we were able to integrate using simple methods such as trigonometric substitution, u-substitution, partial fraction decomposition, and with another algebraic method.

Being me, I didn't think this was enough, and attempted to solve this using a series. Below is my work...

View attachment 240900

My question... Is there a way to continue this series and prove it is equivalent to the algebraic form found using the other integration methods? That is, besides just saying they come from the same integral.

Thank you in advance

Comparison of your answer with $$\ln(1 + y) = \sum_{n=1}^\infty (-1)^{n+1}\frac yn$$ for $|y| < 1$ suggests a problem, and the problem is that you did not start with the correct series expansion of $1/(1 + y)$.

The binomial series expansion $$(1 + y)^{-1} = \sum_{n=0}^\infty (-1)^ny^n$$ converges absolutely when $|y| < 1$. Now $0 < e^x < 1$ only for $x > 0$; for $x < 0$ you will need to use a different expansion.

I leave you to show that:

(1) For $0 \leq a < b$ we have $$\int_a^b \frac{1}{1 + e^x}\,dx = \int_a^b \sum_{n=0}^{\infty} (-1)^ne^{nx}\,dx <br /> = b - a + \sum_{n=1}^\infty (-1)^{n} \left(\frac{e^{nb}}{n} - \frac{e^{na}}{n} \right).$$ (2) This last series converges absolutely (by, eg., the ratio test) and we are therefore justified in reordering the terms.
(3) By reordering the terms we can obtain $$b - a + \sum_{n=1}^\infty (-1)^{n} \left(\frac{e^{nb}}{n} - \frac{e^{na}}{n} \right) = b - a - \ln(e^b + 1) + \ln(e^a + 1) = \left[x - \ln(e^x + 1)\right]_a^b$$ as required.

I also leave you to handle the case $a < b \leq 0$ where this expansion of $1/(1 + e^x)$ does not converge.

(The case $a < 0 < b$ can be obtained from the above two since $\int_a^b = \int_a^0 + \int_0^b$.)

 

[Comeback City]

Comparison of your answer with $$\ln(1 + y) = \sum_{n=1}^\infty (-1)^{n+1}\frac yn$$ for $|y| < 1$ suggests a problem, and the problem is that you did not start with the correct series expansion of $1/(1 + y)$.

The binomial series expansion $$(1 + y)^{-1} = \sum_{n=0}^\infty (-1)^ny^n$$ converges absolutely when $|y| < 1$. Now $0 < e^x < 1$ only for $x > 0$; for $x < 0$ you will need to use a different expansion.

I leave you to show that:

(1) For $0 \leq a < b$ we have $$\int_a^b \frac{1}{1 + e^x}\,dx = \int_a^b \sum_{n=0}^{\infty} (-1)^ne^{nx}\,dx<br /> = b - a + \sum_{n=1}^\infty (-1)^{n} \left(\frac{e^{nb}}{n} - \frac{e^{na}}{n} \right).$$ (2) This last series converges absolutely (by, eg., the ratio test) and we are therefore justified in reordering the terms.
(3) By reordering the terms we can obtain $$b - a + \sum_{n=1}^\infty (-1)^{n} \left(\frac{e^{nb}}{n} - \frac{e^{na}}{n} \right) = b - a - \ln(e^b + 1) + \ln(e^a + 1) = \left[x - \ln(e^x + 1)\right]_a^b$$ as required.

I also leave you to handle the case $a < b \leq 0$ where this expansion of $1/(1 + e^x)$ does not converge.

(The case $a < 0 < b$ can be obtained from the above two since $\int_a^b = \int_a^0 + \int_0^b$.)

Dang, I figured I may have made a mistake early on in the problem... looks like the expansion I attempted to use was a cross between that of (1-x)^-1 and that of e^x itself... wrong in both cases

The part of your explanation I'm having trouble understanding is...

$0 \leq a < b$ we have $$\int_a^b \frac{1}{1 + e^x}\,dx = \int_a^b \sum_{n=0}^{\infty} (-1)^ne^{nx}\,dx<br /> = b - a + \sum_{n=1}^\infty (-1)^{n} \left(\frac{e^{nb}}{n} - \frac{e^{na}}{n} \right).$$

How exactly does the (b-a) come into play at the beginning of the right side of this equation? If I were to take an educated guess, I'd say it has to do with the fact that n=0 must be changed to n=1 as the initial term of the series, along with the definite integral which, as you show, gives the (b-a).

Besides this part, your explanation makes complete sense, thank you!

 

[Comeback City]

Did you justify the change of the integral and the series?

@pasmith found my error, which was simply that I used an incorrect expansion

 

[Comeback City]

@pasmith
I answered my own question... I see that you removed the n=0 term from the series before even taking the integral. Smart thinking. Everything else makes sense, so thank you for your help!

Edit: It looks like in your series expansion of ln(1+y), you forgot that y is raised to the "n" power within the series.