Question

Integral calculus challenge

Answer 1

For a > b > 0, evaluate the integral:

\[\large{\int_{0}^{\infty} \cfrac{e^{ax} - e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx}\]

Answer 2

I would post the complete solution tomorrow :)

For a hint PM me :)

Answer 3

@ganeshie8 @Kainui

Answer 4

@Mimi_x3 is already here :)

Answer 5

@Mimi_x3 This user only accepts messages from people they have fanned. ;)

Answer 6

I can't reply you :)

Answer 7

Ooooh this is m what I'm looking for.

Answer 8

:)

Answer 9

@Mimi_x3 Sent please check if \(\large{\LaTeX}\) is working for you

Answer 10

I repeat-

If anyone wants a hint please message me :)

Answer 11

+-1 in num, separate them 0, substitute x=1/t then use by parts

Answer 12

Looks changing it to double integral is simplifying somewhat

\[\large{\int_{0}^{\infty} \cfrac{e^{ax} - e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx} = \int \limits_0^{\infty} \int \limits _b^a \left(\dfrac{e^{tx}}{x(e^{tx}+1)}\right)'dt~dx\]


\[\large = \int \limits_0^{\infty} \int \limits _b^a \dfrac{e^{tx}}{(e^{tx}+1)^2} dt~dx\]

since the bounds are constants, we can change the order of integration :

\[\large = \int \limits _b^a\int \limits_0^{\infty} \dfrac{e^{tx}}{(e^{tx}+1)^2} dx~dt\]

\[\large = \int \limits _b^a \dfrac{-1}{2t}~dt\]

\[\large = \dfrac{1}{2}\ln (a/b)\]

Answer 13

Correct !! Good work @ganeshie8 :D

Answer 14

That's really cool @ganeshie8 awesome!!!! Wow!

Answer 15

:) more pretty ideas :
http://math.stackexchange.com/questions/590774/improper-integral-of-int-0-infty-frace-ax-e-bxx-dx

Answer 16

For a suitable constant C, set

\[\large{f(x) = \cfrac{e^x}{e^x+1} + c}\]

and show that for t > 0:

\[\large{\int_{0}^{t}\cfrac{(e^{ax}-e^{bx})}{x(e^{ax}+1)(e^{bx}+1)}dx = \int_{0}^{t}\cfrac{f(ax)}{x}dx - \int_{0}^{t}\cfrac{f(bx)}{x}dx}\]

This was the hint and thus I followed this idea :)

Answer 17

i don't remember much of laplace transforms, but im sure it would be much simpler to use below method for our problem also :

http://math.stackexchange.com/questions/294383/evaluate-int-0-infty-left-fracx-textex-texte-x/295326#295326

Answer 18

Good idea though I too haven't read Laplace transformations

Answer 19

I will post my solution tomorrow... feeling too lazy today :P ;)

Answer 20

Interesting, I'm still sort of wondering how you could see how to turn it into a double integral like this in other problems, it doesn't feel very obvious to me yet.

Answer 21

it is by no means obvious lol, i had to delete my reply 20 times while cooking up the first line :P

>>>
```
vishweshshrimali5
Did you reply to the calculus challenge ?

Notification says yes but I can't find any...

Is this a problem from my side ?
```

Answer 22

I guess my best sort of "analytic" approach to this is to just write it as being two identical fractions where I eliminate all the other parts that don't contain the variable like this:\[\Large \frac{e^{ax}}{x(e^{ax}+1)}-\frac{e^{bx}}{x(e^{bx}+1)}\]

Then from there just sort of multiply it back together and hope it works. Although I suppose this is more of a problem of dealing with fractions and not entirely calculus.

Answer 23

nice :) this looks more plausible xD

Answer 24

@ganeshie8
You method is cool..I can't imagine that you have done it so neat.
But I don't understand \[\large{\int_{0}^{\infty} \cfrac{e^{ax} - e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx} = \int \limits_0^{\infty} \int \limits _b^a \left(\dfrac{e^{tx}}{x(e^{tx}+1)}\right)'dt~dx \]
Should it not be
\[ \large{\int_{0}^{\infty} \cfrac{e^{ax} - e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx} = \int \limits_0^{\infty} \int \limits _b^a \left(\dfrac{-1}{x(e^{tx}+1)}\right)'dt~dx \] ?
Am I getting it wrong?

Answer 25

can i use complex ?

Answer 26

@ShailKumar This is wrong, although you could reverse your limits of integration to be correct. It feels odd to do f(a)-f(b) instead of f(b)-f(a) for some reason to me at least, so this might be where your mistake is.

Answer 27

Oh was there something wrong? to me both looks fine still...

in my expression also, numerator simplifies to `e^ax-e^bx` after taking bounds for the expression inside derivative

\[
\large \int \limits _b^a \left(\dfrac{e^{tx}}{x(e^{tx}+1)}\right)'dt = \dfrac{e^{tx}}{x(e^{tx}+1)} \Bigg|_b^a = \cfrac{e^{ax} - e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}
\]

my expression is bit loose, i see it >.<

Answer 28

@ganeshie8 Oh.. I get it now..Thanks

Answer 29

@Kainui Can you be more explicit as I don't understand your point..? :(

Answer 30

My point is that what you had written had a negative sign, so you could fix it by changing your bounds of integration because:

\[\LARGE \int\limits_a^b f(x)dx = \int\limits _b^a -f(x)dx\]

Answer 31

@Kainui When I am integrating from 2 to 1 ?

Answer 32

You're not? I haven't said anything about 2 or 1?

Answer 33

For any constant C, the function

\[\large{f(x) = \cfrac{e^x}{e^x + 1} + c,~~ x\ge 0,}\]

is such that

\[\large{f(ax) - f(bx) = \cfrac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)}}\]

For t > 0 we have:

\[\large{I(t) = \int_{0}^{t}\cfrac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx}\]

\[\large{=\int_{0}^{t}\cfrac{f(ax)-f(bx)}{x}dx}\]

\[\large{=\int_{0}^{t}\cfrac{f(ax)}{x}dx-\int_{0}^{t}\cfrac{f(bx)}{x}dx}\]

Of course, provided both the latter integrals exist.

Now this is true if \(\large{\lim_{y \rightarrow 0} \cfrac{f(y)}{y}}\) exists, which is true iff \(\large{C = -\cfrac{1}{2}}\).

Assuming that this is true, we have :

\[\large{I(t) = \int_{0}^{at}\cfrac{f(y)}{y}dy - \int_{0}^{bt}\cfrac{f(y)}{y}dy}\]

\[\large{=\int_{at}^{bt}\cfrac{f(y)}{y}dy}\]

Now since \(\large{\lim_{x\rightarrow \infty} f(x) = \cfrac{1}{2}}\) so this means that given any \(\large{\epsilon >0}\) there exists a positive real number \(\large{x_{0} = x_{0}(\epsilon)}\)such that

\[\large{x > x_{0}~\implies \cfrac{1}{2} - \epsilon < f(x) < \cfrac{1}{2} + \epsilon}\]

If \(\large{t > \cfrac{x_{0}}{b}}\), then \(\large{at > bt > x_{0}}\), and so we have:

\[\large{(\cfrac{1}{2} - \epsilon)\ln{\cfrac{a}{b}} = \int_{at}^{bt}\cfrac{(\cfrac{1}{2}-\epsilon)}{y}dy}\]

\[\large{< I(t)}\]

\[\large{< \int_{at}^{bt}\cfrac{(\cfrac{1}{2}+\epsilon)}{y}dy}\]

\[\large{= (\cfrac{1}{2}+\epsilon)\ln{\cfrac{a}{b}}}\]

Since, \(\large{\epsilon}\) is arbitrary, thus:

\[\large{\lim_{t\rightarrow \infty} I(t) = \cfrac{1}{2}\ln{\cfrac{a}{b}}}\]

Thus,

\[\large{\int_{0}^{\infty}\cfrac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx = \cfrac{1}{2}\ln{\cfrac{a}{b}}}\]

Answer 34

@ganeshie8
@Kainui

Answer 35

This was my solution ^^^^

Answer 36

cool ^

Answer 37

:) Thanks

Answer 38

I like it, just takes a while to read haha. Very awesome, I'm always looking for new tricks to solve integrals, they are one of my favorite things to play with. Thanks! @vishweshshrimali5

Answer 39

Your welcome :) I too learn new tricks daily thanks to @ganeshie8 and you ;) :D

Answer 40

Awesome @vishweshshrimali5

Answer 41

Thanks @iambatman :)

Answer 42

I started reading this thinking it was new and thought, "Hey, this looks familiar." haha