1. A fun integral

I saw this one and found a cool solution, whats yours?

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

2. Let's start by the typical solution: $
\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: $
\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: $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}
$

$
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 $
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: $
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: $
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 $
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 $u\geq 0$: $
\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: $
\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: $
\left| {\int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}}
{u}} du} \right| \to 0
$

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

3. Originally Posted by PaulRS
Let's start by the typical solution: $
\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: $
\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: $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}
$

$
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 $
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: $
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: $
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 $
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 $u\geq 0$: $
\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: $
\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: $
\left| {\int_0^{ + \infty } {e^{ - u} \cdot \tfrac{{R_n \left( u \right)}}
{u}} du} \right| \to 0
$

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

Let

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

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

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

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

So

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

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

Now seeing that

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

$=\int_0^10~dx$

$=0$

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

$=\ln|1+\alpha|$

So seeing that

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

$=J(1)$

$=\ln|1+1|$

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

So $\int_0^1\frac{x^{99}-1}{\ln(x)}~dx=\ln(100)$

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

$\int_0^1\frac{x-1}{\ln(x)}~dx$
Let $t = \ln x$ then $\int \limits_{-\infty}^0 \frac{e^{2t} - e^t}{t} dt$.
For aesthetic reasons let $x=-t$ to get $\int_0^{\infty} \frac{e^{-x}-e^{-2x}}{x} dx$.

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

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

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

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

Using the Gamma function, $\sum_{n=0}^{\infty} \frac{(-1)^n \Gamma (n+1)}{(n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n n!}{(n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} = \log 2$

5. 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:

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

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

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

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

Now switch em':

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

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

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

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

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

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 $y\neq -1$. 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 $0^{y+1}$.

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

6. 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:

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

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

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

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

Now switch em':

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

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

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

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

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

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 $y\neq -1$. 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 $0^{y+1}$.

I am just a little iffy about this. I reckon if we want it to be 0 we can. That's what I say., yet
You are forgetting the powerfulness of actual numbers opposed to limits.

Supose that $y=-1$

Then we have

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

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

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

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

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 $\lim_{x\to{0^+}}x^0=1$

7. 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.

8. 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!!

9. 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.