# integral challenge

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• Dec 26th 2008, 01:29 PM
galactus
integral challenge
Here's one to give a go if you so desire. We always like a new integral to chew on, huh?. (Cool)(Happy)

Evaluate:

$\displaystyle \int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx$

$\displaystyle a>b>0$

I know, I know, it's easy. Give it a go anyway.(Rock)
• Dec 26th 2008, 02:01 PM
NonCommAlg
Quote:

Originally Posted by galactus
Here's one to give a go if you so desire. We always like a new integral to chew on, huh?. (Cool)(Happy)

Evaluate:

$\displaystyle \int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx$

$\displaystyle a>b>0$

I know, I know, it's easy. Give it a go anyway.(Rock)

your integral = $\displaystyle \int_0^{\infty} \int_b^a \frac{e^{xy}}{(e^{xy} + 1)^2} \ dy \ dx = \cdots = \frac{1}{2}\ln(a/b).$ (Wink)
• Dec 26th 2008, 02:17 PM
galactus
You more than likely went about it easier than I.

By letting $\displaystyle f(x)=\frac{e^{x}}{e^{x}+1}+C$

I left $\displaystyle f(ax)-f(bx)=\frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)}dx$

Then setting $\displaystyle I(u)=\int_{0}^{u}\frac{f(ax)}{x}dx-\int_{0}^{u}\frac{f(bx)}{x}dx$

Assuming they exist of course.

Anyway, it took me longer than it did you, NCA. You are one sharp cat. I always like to peruse your posts for something useful and new to learn.
You've forgotten more than I know.(Bow)(Giggle)
• Dec 26th 2008, 02:22 PM
Jester
Quote:

Originally Posted by galactus
Here's one to give a go if you so desire. We always like a new integral to chew on, huh?. (Cool)(Happy)

Evaluate:

$\displaystyle \int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x(e^{ax}+1)(e^{bx}+1)}dx$

$\displaystyle a>b>0$

I know, I know, it's easy. Give it a go anyway.(Rock)

Another way is to define

$\displaystyle (1)\;\;\;I(m) = \int_0^\infty \frac{dx}{x \; \left( e^{mx} + 1 \right)}$

so your integral is given by $\displaystyle I(b)-I(a)$

If we differentiate (1) wrt m then

$\displaystyle I'(m) = \int_0^\infty \frac{- dx}{ \; \left( e^{mx} + 1 \right)^2}$

then integrate wrt x gives

$\displaystyle I'(m) = \frac{1}{ 2m} \;\;\; \Rightarrow\;\;\; I(m) = \frac{1}{2}\,\ln m + c$

Then, $\displaystyle I(b)-I(a) = \frac{1}{2} \ln \frac{b}{a}$ as given by NonCommAlg (well done btw)
• Dec 26th 2008, 03:22 PM
NonCommAlg
a shorter way is to notice that your integral is a Frullani's Integral if you let $\displaystyle f(x)=\frac{1}{e^x + 1}.$
• Dec 26th 2008, 03:30 PM
galactus
Cool Observation. I must admit, I should have saw that.
• Dec 26th 2008, 03:32 PM
Mathstud28
Besides the above you could see right away that we can discompose our integral as follows

$\displaystyle \frac{x-y}{(x+1)(y+1)}=\frac{1}{y+1}-\frac{1}{x+1}\implies \frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)}=\frac{1}{e^{bx}+1}-\frac{1}{e^{ax}+1}$

So our integral becomes

$\displaystyle \int_0^{\infty}\left\{\frac{1}{x(e^{bx}+1)}-\frac{1}{x(e^{ax}+1)}\right\}dx$

Now using Frullani's Theorem which states that assuming $\displaystyle f'\in\mathcal{C}$ that $\displaystyle \int_0^{\infty}\frac{f(bx)-f(ax)}{x}=\left(f(0)-f(\infty)\right)\ln\left(\frac{a}{b}\right)$

So if we let $\displaystyle f(x)=\frac{1}{e^x+1}$

We get $\displaystyle \int_0^{\infty}\left\{\frac{1}{x(e^{bx}+1)}-\frac{1}{x(e^{ax}+1)}\right\}dx=\left(\frac{1}{e^0 +1}-\frac{1}{e^{\infty}+1}\right)\ln\left(\frac{a}{b}\ right)=\frac{1}{2}\ln\left(\frac{a}{b}\right)$

EDIT: NonCommAlg...you ruin all my fun. ;)
• Dec 26th 2008, 03:33 PM
Jester
I once saw the following on a Putnam exam

$\displaystyle \int_0^{\infty} \frac{\tan^{-1} a x - \tan^{-1} b x }{x}\; dx$

What a wonderful generalization in the link you provide - thanks for sharing! (Clapping)
• Dec 26th 2008, 03:35 PM
Mathstud28
Quote:

Originally Posted by danny arrigo
I once saw the following on a Putnam exam

$\displaystyle \int_0^{\infty} \frac{\tan^{-1} a x - \tan^{-1} b x }{x}\; dx$

What a wonderful generalization in the link you provide - thanks for sharing! (Clapping)

This was in the Putnam :eek:. That is not on the same level as the Putnam integrals I have seen.
• Dec 26th 2008, 03:52 PM
NonCommAlg
Quote:

Originally Posted by Mathstud28
This was in the Putnam :eek:. That is not on the same level as the Putnam integrals I have seen.

the point is that a (good) first-year undergrad student is usually much better at intergration than a second-year student and things get worse and worse as you get a higher education in math.

that's why most graduate students totally suck at it! (Nod)
• Dec 26th 2008, 03:57 PM
Mathstud28
Quote:

Originally Posted by NonCommAlg
the point is that a (good) first-year undergrad student is usually better at intergration than a second-year student and things get worse and worse as you get a higher education in math.

that's why most graduate students totally suck at it! (Nod)

Haha thats true...but still I found some integrals I have seen from past Putnams:

Without CA calculate

$\displaystyle \int_{-\infty}^{\infty}\frac{x^3\sin(x)}{(1+x^2)^2}$

$\displaystyle \int_0^{\infty}\frac{\cos(kx)}{1+x+x^2}dx~~k\geqsl ant 0$

They are a lot of work....especially the first one...if I remember correctly when I did it I used Leibniz's Rule THREE times...(Whew)
• Dec 26th 2008, 04:01 PM
Jester
Quote:

Originally Posted by NonCommAlg
the point is that a (good) first-year undergrad student is usually much better at intergration than a second-year student and things get worse and worse as you get a higher education in math.

that's why most graduate students totally suck at it! (Nod)

Well you know how the saying goes - a mile wide and a foot deep and a mile deep and a foot wide. So, after many years of study, I know a tremendous amount about absolutley nothing. (Nod)
• Dec 26th 2008, 04:06 PM
NonCommAlg
Quote:

Originally Posted by Mathstud28
Haha thats true...but still I found some integrals I have seen from past Putnams:

Without CA calculate

$\displaystyle \int_{-\infty}^{\infty}\frac{x^3\sin(x)}{(1+x^2)^2}$

They are a lot of work....especially the first one...if I remember correctly when I did it I used Leibniz's Rule THREE times...(Whew)

let me guess! the answer is $\displaystyle \frac{\pi}{2e}$ ?
• Dec 26th 2008, 04:10 PM
Mathstud28
Quote:

Originally Posted by NonCommAlg
let me guess! the answer is $\displaystyle \frac{\pi}{2e}$ ?

You mean $\displaystyle \frac{\pi}{2e}$ and yes....but I sense some Complex Analysis in the mix here ;)

EDIT: You corrected yourself.
• Dec 26th 2008, 04:11 PM
Jester
Quote:

Originally Posted by NonCommAlg
let me guess! the answer is $\displaystyle \frac{\pi}{2e}$ ?

Simon Fraser University - I once went to high school with a Dave Muraki - know him?
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