Question

Hello! I'm working on an integral for part of a problem, where I have to solve for \(B\).

It looks something like this:
\(\Large\varepsilon=\int_0^\infty(Be^{cx}\mp1)^{-1}dx\)
This would be so much easier if the integrand was to the first power! Thanks for any help.

Answer 1

Later I will want \(\large e^{-cx}\) still, so that the integral converges.

Answer 2

If it weren't for that -/+ the integral could be done by u-sub, which would lead to partial fractions and it could be done from there. Not sure how any of those methods change when you throw that -/+ in there O.o

Answer 3

Well, I'm just using that to combine two equations for different physical situations. They are
\(\Large\varepsilon=\int_0^\infty(Be^{cx}-1)^{-1}dx\)
and
\(\Large\varepsilon=\int_0^\infty(Be^{cx}+1)^{-1}dx\)

I've been looking into substitutions, but haven't come up with anything helpful yet, I think.

I've considered \(\large u=Be^{cx}+1\), for example. But this leads to... Oh... Hmm... Hehe, I should take another look at that!

Thanks!

Answer 4

Yeah do that u-sub, it'll lead to a partial fractions problem and it comes up very easily.

Answer 5

I'll give that a shot...

Answer 6

Turning out okay? :)

Answer 7

\(\large{ u=Be^{cx}\mp1\\\quad\Rightarrow x=\dfrac{\ln\left(\dfrac{u\pm1}B\right)}c\\\quad\Rightarrow \text du=Bce^{cx}\text dx\\\quad\quad\Rightarrow dx=\left(Bce^{cx}\right)^{-1}du}\)

This looks ugly so far..

Answer 8

Nuu, no need to do that substitution

Answer 9

Ill start ya out :P

Answer 10

For \(x\)?

\(\text dx\) depends on \(x\) though, right?

I'll watch what you do.

Answer 11

\[u = Be^{cx} + 1\]
\[\frac{ u-1 }{ B } = e^{cx}\]
\[du = Bce^{cx}dx\]
\[dx = \frac{ du }{ Bce^{cx} }\]
All of this gives us:

\[\int\limits_{0}^{\infty}\frac{ 1 }{ Bce^{cx}*u }du = \int\limits_{0}^{\infty}\frac{ 1 }{ Bc \frac{ u-1 }{ B } *u}du = \frac{ 1 }{c } \int\limits_{0}^{\infty}\frac{ 1 }{ u(u-1) }du\]

So the key was to solve for e^(cx), not x itself.

Answer 12

Oh, thank you very much! And now you suggest partial fractions, right? Otherwise I would tackle this with integration by parts just because I'm more familiar with that. But I have been taught to use partial fractions.

Answer 13

Yeah, partial fractions seems more natural to me. I see factoring on bottom, so that keys to me partials. If you know how to go about that, it should come out pretty well

Answer 14

Okay, thanks! I'll work on that, then.

Answer 15

Alright, cool. I gotta head out now, so good luck :) Glad that helps.

Answer 16

Thank you for your help! Take care! :)

Answer 17

We left off with
\[\large{\dfrac1c\int_0^\infty \dfrac1{u(u\pm1)}du}\]
This becomes
\[\large{\dfrac1c}\int_0^\infty \left(\dfrac {A(u\pm1)+Bu}{u(u\pm1)}\right)du\]\(\qquad\)Note, \(1=A(u\pm1)+Bu\)
\(\qquad\qquad\)So, \(u=0\Rightarrow1=\pm A\qquad\Rightarrow\qquad A=\mp1\)
\(\qquad\qquad\) and \(u=\mp1\Rightarrow1=\mp B\qquad\Rightarrow\qquad B=\pm1\)

\[\large{\dfrac1c}\int_0^\infty \left(\dfrac Au+\dfrac B{u\pm1}\right)du\]

Putting this together, we have
\[\large{\dfrac1c}\int_0^\infty \left(\dfrac {\mp1}u+\dfrac {\pm1}{u\pm1}\right)du\]

So, there is the partial fractions part.

Answer 18

I made a sign error when determining \(A\) and \(B\).
\(\large u=0\Rightarrow1=\pm A\qquad\Rightarrow\qquad A=\pm1\)
\(u=\mp1\Rightarrow1=\mp B\qquad\Rightarrow\qquad B=\mp1\)
There was no justification for this mistake.

We can break the integral up to be
\[\large{\dfrac1c}\left(\int_0^\infty \dfrac {\pm1}u\text du+\int_0^\infty\dfrac {\mp1}{u\pm1}\text du\right)\]

It can be made to look nicer, like
\[\large{\dfrac1c}\left(\pm\int_0^\infty u^{-1}\text du\mp\int_0^\infty(u\pm1)^{-1}\text du\right)\]

This becomes
\[\large{\dfrac1c}\left(\pm\left[\ln(u)\right]_0^\infty\mp\left[\ln(u\pm1)\right]_0^\infty\right)\]

Now we go back to \(\large u=Be^{cx}\mp1\). With substitution, we have
\[\large{\dfrac1c}\left(\pm\left[\ln(Be^{cx}\mp1)\right]_0^\infty\mp\left[\ln(Be^{cx}\mp1\pm1)\right]_0^\infty\right)\]which simplifies to be
\[\large{\dfrac1c}\left(\pm\left[\ln(Be^{cx}\mp1)\right]_0^\infty\mp\left[\ln(Be^{cx})\right]_0^\infty\right)\]

For clarity, the last expression captures two:
\[\large{\dfrac1c}\left(+\left[\ln(Be^{cx}-1)\right]_0^\infty-\left[\ln(Be^{cx})\right]_0^\infty\right)\]and
\[\large{\dfrac1c}\left(-\left[\ln(Be^{cx}+1)\right]_0^\infty+\left[\ln(Be^{cx})\right]_0^\infty\right)\]

Answer 19

Another mistake I noticed is that I left the bounds the same when integrating with respect to \(u\). I should've noted that they were bounds with respect to \(x\), or omitted them altogether.

I think that I can simpify like this:\[\large{\dfrac1c}\left(\pm\left[\ln(Be^{cx}\mp1)\mp\ln(Be^{cx})\right]_0^\infty\right)\]

This allows the natural log functions to be multiplied or divided (depending on sign),
\[\large{\dfrac1c}\left(\pm\left[\ln\left((Be^{cx}\mp1)(Be^{cx})^{\mp1}\right)\right]_0^\infty\right)\]

Now it will be better to work with two different equations.
Distributing with the top operator we get
\[\large{\dfrac1c}\left(+\left[\ln\left((Be^{cx}-1)(Be^{cx})^{-1}\right)\right]_0^\infty\right)\\~\\=\large{\dfrac1c}\left(\left[\ln\left(1-(Be^{cx})^{-1}\right)\right]_0^\infty\right)
\\~\\=\large{\dfrac1c}\left(\left[\ln\left(1-B^{-1}e^{-cx}\right)\right]_0^\infty\right)
\\~\\=\large{\dfrac1c\left(\ln(1)-\ln(1-B^{-1})\right)}
\\~\\=\large{\dfrac1c\ln\left(\dfrac1{1-\dfrac1B}\right)=\varepsilon}
\]

Okay!
\[\large{\ln\left(\dfrac1{1-\dfrac1B}\right)=c\varepsilon
\\~\\\Rightarrow \dfrac1{1-\dfrac1B}=e^{c\varepsilon}
\\~\\\Rightarrow 1-\dfrac1B=e^{-c\varepsilon}
\\~\\\Rightarrow\dfrac1B=1-e^{-c\varepsilon}
\\~\\\Rightarrow B=(1-e^{-c\varepsilon})^{-1}}\]

And, if I go through with the bottom operations, it should come out to \(B=-(1-e^{-c\varepsilon})^{-1}\).

Answer 20

For the sake of anyone reading this, it seems that I made a mistake. There should be no exponent of \(\mp1\), it should just be \(-1\).

Changing that, I got to the correct result. Also, my book provides a formula for this that differs in that exponent only. I haven't found my mistake, but I might search for it another time.