# A fun integral

• Jul 17th 2008, 01:59 PM
Mathstud28
A fun integral
I saw this one and found a cool solution, whats yours?

$\displaystyle \int_0^1\frac{x-1}{\ln(x)}~dx$
• Jul 17th 2008, 02:46 PM
PaulRS
Let's start by the typical solution: $\displaystyle \int_0^1 {\tfrac{{x - 1}} {{\ln \left( x \right)}}dx} = \int_0^1 {\left( {\int_0^1 {x^u du} } \right)dx = } \int_0^1 {\left( {\int_0^1 {x^u dx} } \right)du}$

Thus: $\displaystyle \int_0^1 {\tfrac{{x - 1}} {{\ln \left( x \right)}}dx} = \int_0^1 {\tfrac{{du}} {{u + 1}}} = \ln \left( 2 \right)$

We also have: $\displaystyle I= \int_0^1 {\tfrac{{x - 1}} {{\ln \left( x \right)}}dx} \underbrace = _{u = - \ln \left( x \right)} - \int_0^{ + \infty } {\left( {e^{ - u} } \right) \cdot \left( {\tfrac{{e^{ - u} - 1}} {u}} \right)du}$

$\displaystyle I = - \int_0^{ + \infty } {\left( {e^{ - u} } \right) \cdot \left( {\tfrac{{ - u + \tfrac{{u^2 }} {2} - \tfrac{{u^3 }} {{3!}} \pm ... + \tfrac{{\left( { - 1} \right)^n \cdot u^n }} {{n!}} + R_n \left( u \right)}} {u}} \right)du}$ then $\displaystyle I = \int_0^{ + \infty } {\left( {e^{ - u} } \right) \cdot \left( {1 - \tfrac{u} {2} + \tfrac{{u^2 }} {{3!}} \mp ... - \tfrac{{\left( { - 1} \right)^n \cdot u^{n - 1} }} {{n!}} - \tfrac{{R_n \left( u \right)}} {u}} \right)du}$

Since each of those integrals exists separately: $\displaystyle I = \int_0^{ + \infty } {e^{ - u} d} u - \tfrac{1} {2}\int_0^{ + \infty } {u \cdot e^{ - u} du} ... + \tfrac{{\left( { - 1} \right)^{n + 1} }} {{n!}} \cdot \int_0^{ + \infty } {u^{n - 1} \cdot e^{ - u} du} - \int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du$

We have: $\displaystyle I = 1 - \tfrac{1} {2} + {\kern 1pt} \tfrac{1} {3} \mp ... + \tfrac{{\left( { - 1} \right)^{n + 1} }} {n} - \int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du$ where the Remainder is $\displaystyle R_n \left( u \right) = \tfrac{{\left( { - 1} \right)^{n + 1} }} {{n!}} \cdot \int_0^u {\left( {u - t} \right)^n \cdot e^{ - t} } dt$ (see the Remainder of Taylor's Polynomials)

Since $\displaystyle u\geq 0$: $\displaystyle \left| {R_n \left( u \right)} \right| = \left| {\tfrac{{\left( { - 1} \right)^{n + 1} }} {{n!}} \cdot \int_0^u {\left( {u - t} \right)^n \cdot e^{ - t} } dt} \right| \leqslant \tfrac{{u^n }} {{n!}} \cdot \int_0^u {e^{ - t} } dt \leqslant \tfrac{{u^n }} {{n!}}$ thus: $\displaystyle \left| {\int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du} \right| \leqslant \int_0^{ + \infty } {\left| {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} \right|} du \leqslant \tfrac{1} {{n!}} \cdot \int_0^{ + \infty } {e^{ - u} \cdot u^{n - 1} } du = \tfrac{1} {n}$

Therefore the extra term disappears as n tends to infinity: $\displaystyle \left| {\int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du} \right| \to 0$

Thus: $\displaystyle I = 1 - \tfrac{1} {2} + \tfrac{1} {3} \mp ... = \ln \left( 2 \right)$
• Jul 17th 2008, 02:56 PM
Mathstud28
Quote:

Originally Posted by PaulRS
Let's start by the typical solution: $\displaystyle \int_0^1 {\tfrac{{x - 1}} {{\ln \left( x \right)}}dx} = \int_0^1 {\left( {\int_0^1 {x^u du} } \right)dx = } \int_0^1 {\left( {\int_0^1 {x^u dx} } \right)du}$

Thus: $\displaystyle \int_0^1 {\tfrac{{x - 1}} {{\ln \left( x \right)}}dx} = \int_0^1 {\tfrac{{du}} {{u + 1}}} = \ln \left( 2 \right)$

We also have: $\displaystyle I= \int_0^1 {\tfrac{{x - 1}} {{\ln \left( x \right)}}dx} \underbrace = _{u = - \ln \left( x \right)} - \int_0^{ + \infty } {\left( {e^{ - u} } \right) \cdot \left( {\tfrac{{e^{ - u} - 1}} {u}} \right)du}$

$\displaystyle I = - \int_0^{ + \infty } {\left( {e^{ - u} } \right) \cdot \left( {\tfrac{{ - u + \tfrac{{u^2 }} {2} - \tfrac{{u^3 }} {{3!}} \pm ... + \tfrac{{\left( { - 1} \right)^n \cdot u^n }} {{n!}} + R_n \left( u \right)}} {u}} \right)du}$ then $\displaystyle I = \int_0^{ + \infty } {\left( {e^{ - u} } \right) \cdot \left( {1 - \tfrac{u} {2} + \tfrac{{u^2 }} {{3!}} \mp ... - \tfrac{{\left( { - 1} \right)^n \cdot u^{n - 1} }} {{n!}} - \tfrac{{R_n \left( u \right)}} {u}} \right)du}$

Since each of those integrals exists separately: $\displaystyle I = \int_0^{ + \infty } {e^{ - u} d} u - \tfrac{1} {2}\int_0^{ + \infty } {u \cdot e^{ - u} du} ... + \tfrac{{\left( { - 1} \right)^{n + 1} }} {{n!}} \cdot \int_0^{ + \infty } {u^{n - 1} \cdot e^{ - u} du} - \int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du$

We have: $\displaystyle I = 1 - \tfrac{1} {2} + {\kern 1pt} \tfrac{1} {3} \mp ... + \tfrac{{\left( { - 1} \right)^{n + 1} }} {n} - \int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du$ where the Remainder is $\displaystyle R_n \left( u \right) = \tfrac{{\left( { - 1} \right)^{n + 1} }} {{n!}} \cdot \int_0^u {\left( {u - t} \right)^n \cdot e^{ - t} } dt$ (see the Remainder of Taylor's Polynomials)

Since $\displaystyle u\geq 0$: $\displaystyle \left| {R_n \left( u \right)} \right| = \left| {\tfrac{{\left( { - 1} \right)^{n + 1} }} {{n!}} \cdot \int_0^u {\left( {u - t} \right)^n \cdot e^{ - t} } dt} \right| \leqslant \tfrac{{u^n }} {{n!}} \cdot \int_0^u {e^{ - t} } dt \leqslant \tfrac{{u^n }} {{n!}}$ thus: $\displaystyle \left| {\int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du} \right| \leqslant \int_0^{ + \infty } {\left| {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} \right|} du \leqslant \tfrac{1} {{n!}} \cdot \int_0^{ + \infty } {e^{ - u} \cdot u^{n - 1} } du = \tfrac{1} {n}$

Therefore the extra term disappears as n tends to infinity: $\displaystyle \left| {\int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}} {u}} du} \right| \to 0$

Thus: $\displaystyle I = 1 - \tfrac{1} {2} + \tfrac{1} {3} \mp ... = \ln \left( 2 \right)$

Nice Solution Paul!

Let

$\displaystyle J(\theta)=\int_0^{1}\frac{x^{\theta}-1}{\ln(x)}~dx$

$\displaystyle \Rightarrow{J'(\theta)=\int_0^1\frac{\partial}{\pa rtial\theta}\bigg[\frac{x^{\theta}}{\ln(x)}-\frac{1}{\ln(x)}\bigg]~dx}$

$\displaystyle =\int_0^{1}x^{\theta}~dx$

$\displaystyle =\frac{1}{\theta+1}$

So

$\displaystyle \int_0^{\alpha}J'(\theta)~d\theta$

$\displaystyle =J(\alpha)-J(0)$

Now seeing that

$\displaystyle J(0)=\int_0^1\frac{x^0-1}{\ln(x)}~dx$

$\displaystyle =\int_0^10~dx$

$\displaystyle =0$

$\displaystyle \Rightarrow{J(\alpha)=\int_0^{1}\frac{x^{\alpha}-1}{\ln(x)}=\int_0^1\frac{d\alpha}{1+\alpha}}$

$\displaystyle =\ln|1+\alpha|$

So seeing that

$\displaystyle \int_0^1\frac{x-1}{\ln(x)}~dx$

$\displaystyle =J(1)$

$\displaystyle =\ln|1+1|$

$\displaystyle =\ln(2)\quad\blacksquare$

So $\displaystyle \int_0^1\frac{x^{99}-1}{\ln(x)}~dx=\ln(100)$
• Jul 17th 2008, 03:07 PM
ThePerfectHacker
Quote:

Originally Posted by Mathstud28
I saw this one and found a cool solution, whats yours?

$\displaystyle \int_0^1\frac{x-1}{\ln(x)}~dx$

Let $\displaystyle t = \ln x$ then $\displaystyle \int \limits_{-\infty}^0 \frac{e^{2t} - e^t}{t} dt$.
For aesthetic reasons let $\displaystyle x=-t$ to get $\displaystyle \int_0^{\infty} \frac{e^{-x}-e^{-2x}}{x} dx$.

Now write, $\displaystyle \int_0^{\infty} \int_0^{\infty} (e^{-x} - e^{-2x}) e^{-yx} dy~ dx = \int_0^{\infty}\int_0^{\infty} e^{-x(y+1)} -e^{-x(y+2)}dx ~ dy$

Perform the integration, $\displaystyle \int_0^{\infty} \frac{1}{y+1} - \frac{1}{y+2} dy = \lim_{X\to \infty} \left( \log 2 - \log \frac{X+1}{X+2}\right) = \log 2$

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Here is a different variation on what Paul did. (It is not as rigorous though).

We arrive at, $\displaystyle \int_0^{\infty} e^{-x} \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{(n+1)!} \right) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)!}\left( \int_0^{\infty}e^{-x} x^n dx \right)$.

Using the Gamma function, $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n \Gamma (n+1)}{(n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n n!}{(n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} = \log 2$
• Jul 18th 2008, 07:34 AM
galactus
I may as well give my two cents. I will use the double integral we're all so fond of. I looked around to see if the Kriz may have done this at some time or another, but I couldn't find anything, so I assume not.

An identity we can use is:

$\displaystyle \frac{x-1}{ln(x)}dx=\int_{0}^{1}x^{y}dy$

So, we can use our double integral thing like so:

$\displaystyle \int_{0}^{1}\frac{x-1}{ln(x)}dx$

$\displaystyle \int_{0}^{1}\int_{0}^{1}x^{y}dydx$

Now switch em':

$\displaystyle \int_{0}^{1}\int_{0}^{1}x^{y}dxdy$

But, $\displaystyle \int_{0}^{1}x^{y}dx=\frac{1}{y+1}$

And, $\displaystyle \int_{0}^{1}\frac{1}{y+1}dy=ln(2)$

You know, there is a little something that is bothering me. Please straighten me out if need be.

$\displaystyle \int_{0}^{1}x^{y}dx=\frac{1-\lim_{x\to 0^{+}}x^{y+1}}{y+1}$

While ding this I noticed something. Now, assuming the limit above is 0 we are in good shape, but isn't it undefined?. I supose as long as $\displaystyle y\neq -1$. Maybe I am wrong, but something about that is bothering me an little.

As a check I ran it through my 92 and it said undefined. Yet we should get $\displaystyle 0^{y+1}$.

I am just a little iffy about this. I reckon if we want it to be 0 we can. That's what I say.(Worried), yet (Happy)
• Jul 18th 2008, 01:13 PM
Mathstud28
Quote:

Originally Posted by galactus
I may as well give my two cents. I will use the double integral we're all so fond of. I looked around to see if the Kriz may have done this at some time or another, but I couldn't find anything, so I assume not.

An identity we can use is:

$\displaystyle \frac{x-1}{ln(x)}dx=\int_{0}^{1}x^{y}dy$

So, we can use our double integral thing like so:

$\displaystyle \int_{0}^{1}\frac{x-1}{ln(x)}dx$

$\displaystyle \int_{0}^{1}\int_{0}^{1}x^{y}dydx$

Now switch em':

$\displaystyle \int_{0}^{1}\int_{0}^{1}x^{y}dxdy$

But, $\displaystyle \int_{0}^{1}x^{y}dx=\frac{1}{y+1}$

And, $\displaystyle \int_{0}^{1}\frac{1}{y+1}dy=ln(2)$

You know, there is a little something that is bothering me. Please straighten me out if need be.

$\displaystyle \int_{0}^{1}x^{y}dx=\frac{1-\lim_{x\to 0^{+}}x^{y+1}}{y+1}$

While ding this I noticed something. Now, assuming the limit above is 0 we are in good shape, but isn't it undefined?. I supose as long as $\displaystyle y\neq -1$. Maybe I am wrong, but something about that is bothering me an little.

As a check I ran it through my 92 and it said undefined. Yet we should get $\displaystyle 0^{y+1}$.

I am just a little iffy about this. I reckon if we want it to be 0 we can. That's what I say.(Worried), yet (Happy)

You are forgetting the powerfulness of actual numbers opposed to limits.

Supose that $\displaystyle y=-1$

Then we have

$\displaystyle \frac{1-\lim_{x\to{0^+}}x^0}{y+1}$

Now if we had that $\displaystyle \lim_{x\to{0}}f(x)^{g(x)}$

Where $\displaystyle f(0)=g(0)=0$ that would be indeterminate but we have

$\displaystyle \lim_{x\to{0^+}}f(x)^0$

Now this limit is still a number...a non zero number albeit almost zero so we know that no matter how small it gets it never actually reaches zero...therefore $\displaystyle \lim_{x\to{0^+}}x^0=1$
• Jul 18th 2008, 01:18 PM
galactus
You are so right. I just wanted someones input. I was thinking too much and had a brain cramp, I reckon. Anyway, Pretty cool little problem, mathstud.

It's fun seeing the different ways yo go about it.
• Jul 18th 2008, 01:20 PM
Mathstud28
Quote:

Originally Posted by galactus
You are so right. I just wanted someones input. I was thinking too much, I reckon. Anyway, Pretty cool little problem, mathstud.

It's fun seeing the different ways yo go about it.

That's what I love about math! So much room for individuality! And I know you knew that...we all just want to double check sometimes!! (Sun)
• Jul 18th 2008, 02:35 PM
Krizalid
Quote:

Originally Posted by galactus

I looked around to see if the Kriz may have done this at some time or another, but I couldn't find anything, so I assume not.

Here.