# cool integral

• July 6th 2008, 06:15 AM
galactus
cool integral
Perhaps a number of you are familiar with the solution to this integral. if not, see if you can work it out.

$\int_{0}^{1}\left[\frac{1}{ln(x)}+\frac{1}{1-x}\right]dx$

If you do them individually, you just get ${-\infty}, \;\ {\infty}$

But that is certainly not the solution.

It has the same solution as:

$\int_{0}^{1}\frac{1-e^{-x}-e^{\frac{-1}{x}}}{x}dx$
• October 25th 2008, 08:33 PM
Mathstud28
Quote:

Originally Posted by galactus
Perhaps a number of you are familiar with the solution to this integral. if not, see if you can work it out.

$\int_{0}^{1}\left[\frac{1}{ln(x)}+\frac{1}{1-x}\right]dx$

If you do them individually, you just get ${-\infty}, \;\ {\infty}$

But that is certainly not the solution.

It has the same solution as:

$\int_{0}^{1}\frac{1-e^{-x}-e^{\frac{-1}{x}}}{x}dx$

I have worked on this integral for about two weeks and have finally come up with a solution

First split it into

$\int_0^1\frac{1-e^{-x}}{x}dx-\int_0^1\frac{e^{\frac{-1}{x}}}{x}dx$

Then make the sub $x=\frac{1}{u}$ on the second integral turning it into

$\int_0^1\frac{1-e^{-x}}{x}dx-\int_1^{\infty}\frac{e^{-x}}{x}$

Integration by parts on both gives

$=\int_0^1e^{-x}\ln(x)dx+\int_1^{\infty}e^{-x}\ln(x)dx=\int_0^{\infty}e^{-x}\ln(x)dx$

Now consider the gamma function

$\Gamma(t)=\int_0^{\infty}x^{t-1}e^{-x}dx$

Well since the Gamma function is uniformly convergent for all values of x that make it converge we have that

$\Gamma'(t)=\frac{d}{dx}\int_0^{\infty}x^{t-1}e^{-x}dx=\int_0^{\infty}\frac{\partial}{\partial{t}}x^ {t-1}e^{-x}dx=\int_0^{\infty}x^{t-1}e^{-x}\ln(x)dx$

$\therefore\quad\Gamma'(1)=\int_0^{\infty}e^{-x}\ln(x)dx=-\gamma$

The value of Gamma' evaluated at one...I cannot prove..haha
• October 26th 2008, 05:34 AM
galactus
Here is a tidbit you may find nice:

The DiGamma function:

${\Psi}(1)=\frac{{\Gamma}(1)'}{{\Gamma}(1)}=-1-{\gamma}+\sum_{r=1}^{\infty}\left(\frac{1}{r}-\frac{1}{r+1}\right)=-{\gamma}$
• October 26th 2008, 09:54 AM
Mathstud28
Quote:

Originally Posted by galactus
Here is a tidbit you may find nice:

The DiGamma function:

${\Psi}(1)=\frac{{\Gamma}(1)'}{{\Gamma}(1)}=-1-{\gamma}+\sum_{r=1}^{\infty}\left(\frac{1}{r}-\frac{1}{r+1}\right)=-{\gamma}$

Thanks Galctus...I have heard of the Digamma function but I was unaware of its series representation...I just knew that $\Gamma'(1)=-\gamma$
• December 28th 2008, 09:27 AM
PaulRS
Lemma 1. Let $
f:\mathbb{R}/\mathbb{R}^ - \to \mathbb{R}
$
be a function with continuous derivative in its domain.

Then $
\sum\limits_{k = 1}^n {f\left( k \right)} = \int_0^n {f\left( x \right)dx} + \int_0^n {\left\{ x \right\} \cdot f'\left( x \right)dx}
$
where $
\left\{ x \right\} = x - \left\lfloor x \right\rfloor
$

Proof

$
\int_0^n {\left\{ x \right\} \cdot f'\left( x \right)dx} = \sum\limits_{k = 0}^{n - 1} {\int_k^{k + 1} {\left\{ x \right\} \cdot f'\left( x \right)dx} }
$
$
= \sum\limits_{k = 0}^{n - 1} {\int_k^{k + 1} {\left( {x - k} \right) \cdot f'\left( x \right)dx} }
$
$
= \int_0^n {x \cdot f'\left( x \right)dx} - \sum\limits_{k = 0}^{n - 1} {k \cdot \int_k^{k + 1} {f'\left( x \right)dx} }
$

By the Fundamental Theorem of Calculus: $
\sum\limits_{k = 0}^{n - 1} {k \cdot \int_k^{k + 1} {f'\left( x \right)dx} } = \sum\limits_{k = 0}^{n - 1} {k \cdot \left[ {f\left( {k + 1} \right) - f\left( k \right)} \right]}
$

So: $
\sum\limits_{k = 0}^{n - 1} {k \cdot \left[ {f\left( {k + 1} \right) - f\left( k \right)} \right]} = \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( {k + 1} \right)} - \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( k \right)}
$
$
= \sum\limits_{k = 0}^{n - 1} {\left( {k + 1 - 1} \right) \cdot f\left( {k + 1} \right)} - \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( k \right)}
$
$
= \sum\limits_{k = 0}^{n - 1} {\left( {k + 1} \right) \cdot f\left( {k + 1} \right)} - \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( k \right)} - \sum\limits_{k = 0}^{n - 1} {f\left( {k + 1} \right)}
$
$
= n \cdot f\left( n \right) - \sum\limits_{k = 1}^n {f\left( k \right)}
$

Now integrating by parts: $
\int_0^n {x \cdot f'\left( x \right)dx} = n \cdot f\left( n \right) - \int_0^n {f\left( x \right)dx}
$
and Lemma 1 follows. $
\square
$

Lemma 2. We have: $
\mathop {\lim }\limits_{s \to 1^ + } \left( {\zeta \left( s \right) - \tfrac{1}
{{s - 1}}} \right) = \gamma
$

Proof

First take $
f\left( x \right) = \tfrac{1}
{{x + 1}}
$
in Lemma 1: We get $
\sum\limits_{k = 1}^n {\tfrac{1}
{{k + 1}}} = \log \left( {n + 1} \right) - \int_0^n {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^2 }}dx}
$

Thus: $
\sum\limits_{k = 1}^{n + 1} {\tfrac{1}
{k}} = \log \left( {n + 1} \right) + 1 - \int_0^n {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^2 }}dx}
$
Now note that the integral in the RHS converges as $
n \to + \infty
$
- this in fact shows the existance of the gamma constant- then by definition of the gamma constant: $
\gamma = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sum\limits_{k = 1}^{n + 1} {\tfrac{1}
{k}} - \log \left( {n + 1} \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left( {1 - \int_0^n {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^2 }}dx} } \right) = 1 - \int_0^\infty {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^2 }}dx}
$

Now take $
f\left( x \right) = \tfrac{1}
{{\left( {x + 1} \right)^s }}
$
with $s>1$ in Lemma 1 : $
\sum\limits_{k = 1}^n {\tfrac{1}
{{\left( {k + 1} \right)^s }}} = \tfrac{{\left( {n + 1} \right)^{1 - s} }}
{{1 - s}} - \tfrac{1}
{{1 - s}} - s \cdot \int_0^n {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^{s + 1} }}dx}
$

Thus: $
\sum\limits_{k = 1}^\infty {\tfrac{1}
{{\left( {k + 1} \right)^s }}} = \tfrac{1}
{{s - 1}} - s \cdot \int_0^\infty {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^{s + 1} }}dx}
$
and then $
\zeta \left( s \right) = \tfrac{1}
{{s - 1}} + 1 - s \cdot \int_0^\infty {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^{s + 1} }}dx}
$

Therefore: $
\mathop {\lim }\limits_{s \to 1^ + } \left( {\zeta \left( s \right) - \tfrac{1}
{{s - 1}}} \right) = 1 - \mathop {\lim }\limits_{s \to 1^ + } s \cdot \int_0^\infty {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^{s + 1} }}dx} = 1 - \int_0^\infty {\tfrac{{\left\{ x \right\}}}
{{\left( {x + 1} \right)^2 }}dx} = \gamma
$
$
\square
$

Lemma 3. $
\zeta \left( s \right) \cdot \Gamma \left( s \right) = \int_0^\infty {\tfrac{{x^{s - 1} }}
{{e^x - 1}}dx}
$
: $s>1$
Proof: $
\int_0^\infty {\tfrac{{x^{s - 1} }}
{{e^x - 1}}dx} = \int_0^\infty {x^{s - 1} \cdot \tfrac{{e^{ - x} }}
{{1 - e^{ - x} }}dx} = \int_0^\infty {x^{s - 1} \cdot \sum\limits_{n = 1}^\infty {e^{ - nx} } dx} = \sum\limits_{n = 1}^\infty {\int_0^\infty {x^{s - 1} \cdot e^{ - nx} dx} }
$
now since $
\int_0^\infty {x^{s - 1} \cdot e^{ - nx} dx} = \tfrac{{\Gamma \left( s \right)}}
{{n^s }}
$
we are done $
\square
$

Proof : $
\int_0^1 {\left( {\tfrac{1}
{{1 - x}} + \tfrac{1}
{{\log \left( x \right)}}} \right)dx} = \gamma
$

Let $I$ be the integral then letting $
u = - \log \left( x \right)
$
we get $
I = \int_0^\infty {e^{ - u} \cdot \left( {\tfrac{1}
{{1 - e^{ - u} }} - \tfrac{1}
{u}} \right)du}
$

Now note that: $
\left( {\zeta \left( s \right) - \tfrac{1}
{{s - 1}}} \right) \cdot \Gamma \left( s \right) = \zeta \left( s \right) \cdot \Gamma \left( s \right) - \tfrac{{\Gamma \left( s \right)}}
{{s - 1}} = \zeta \left( s \right) \cdot \Gamma \left( s \right) - \tfrac{{\left( {s - 1} \right) \cdot \Gamma \left( {s - 1} \right)}}
{{s - 1}}
$
( $s>1$ )

Thus: $
\left( {\zeta \left( s \right) - \tfrac{1}
{{s - 1}}} \right) \cdot \Gamma \left( s \right) = \zeta \left( s \right) \cdot \Gamma \left( s \right) - \Gamma \left( {s - 1} \right) = \int_0^\infty {\tfrac{{u^{s - 1} }}
{{e^u - 1}}}du - \int_0^\infty {u^{s - 2} \cdot e^{ - u} du}
$

Then: $
\left( {\zeta \left( s \right) - \tfrac{1}
{{s - 1}}} \right) \cdot \Gamma \left( s \right) = \int_0^\infty {\left( {\tfrac{{u^{s - 1} }}
{{e^u - 1}} - u^{s - 2} \cdot e^{ - u} } \right)} du = \int_0^\infty {u^{s - 1} e^{ - u} \cdot \left( {\tfrac{1}
{{1 - e^{ - u} }} - \tfrac{{1 }}
{u}} \right)} du
$

Applying Lemma 2: $
\gamma = \mathop {\lim }\limits_{s \to 1^ + } \left( {\zeta \left( s \right) - \tfrac{1}
{{s - 1}}} \right) \cdot \Gamma \left( s \right) = \mathop {\lim }\limits_{s \to 1^ + } \int_0^\infty {u^{s - 1} e^{ - u} \cdot \left( {\tfrac{1}
{{1 - e^{ - u} }} - \tfrac{1}
{u}} \right)} du
$
$
= \int_0^\infty {e^{ - u} \cdot \left( {\tfrac{1}
{{1 - e^{ - u} }} - \tfrac{{1 }}
{u}} \right)} du \square
$
• December 29th 2008, 12:07 PM
PaulRS
Lemma. $
\sum\limits_{k = 2}^\infty {\tfrac{{\zeta \left( k \right)}}
{k} \cdot \left( { - 1} \right)^k } = \gamma
$

Proof

By definition: $
\gamma = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sum\limits_{k = 1}^n {\left( {\tfrac{1}
{k}} \right)} - \log \left( {n + 1} \right)} \right]
$

Note that: $
\log \left( {n + 1} \right) = \sum\limits_{k = 1}^n {\log \left( {1 + \tfrac{1}
{k}} \right)}
$
thus: $
\sum\limits_{k = 1}^n {\left( {\tfrac{1}
{k}} \right)} - \log \left( {n + 1} \right) = \sum\limits_{k = 1}^n {\left[ {\tfrac{1}
{k} - \log \left( {1 + \tfrac{1}
{k}} \right)} \right]}
$

We get: $
\gamma = \sum\limits_{k = 1}^\infty {\left[ {\tfrac{1}
{k} - \log \left( {1 + \tfrac{1}
{k}} \right)} \right]}
$
Consider that: $
\log \left( {1 + \tfrac{1}
{k}} \right) = \tfrac{1}
{k} - \tfrac{1}
{{2 \cdot k^2 }} \pm ...
$
then $
\gamma = \sum\limits_{k = 1}^\infty {\sum\limits_{s = 2}^\infty {\tfrac{{\left( { - 1} \right)^s }}
{{s \cdot k^s }}} }
$

Exchange the summation order ( this can be justified working with the remainder, see here): $
\gamma = \sum\limits_{s = 2}^\infty {\tfrac{{\zeta \left( s \right)}}
{s} \cdot \left( { - 1} \right)^s } \blacksquare
$

Proof. $\int_{0}^{1}\left(\frac{1}{\log(x)}+\frac{1}{1-x}\right)dx=\gamma
$

Letting $
{u = - \log \left( x \right)}
$
we have: $
\int_0^1 {\left( {\tfrac{1}
{{\log \left( x \right)}} + \tfrac{1}
{{1 - x}}} \right)dx} = \int_0^{ + \infty } {e^{ - u} \cdot \left( { - \tfrac{1}
{u} + \tfrac{1}
{{1 - e^{ - u} }}} \right)du} =
\int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
{{u \cdot \left( {e^u - 1} \right)}} \cdot du}

$

Now use the Power series expansion of $e^x$: $
\int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
{{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \int_0^{ + \infty } {\tfrac{{\tfrac{{u^2 }}
{{2!}} - \tfrac{{u^3 }}
{{3!}} + \tfrac{{u^4 }}
{{4!}} \mp ...}}
{{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \int_0^{ + \infty } {\tfrac{{\tfrac{u}
{{2!}} - \tfrac{{u^2 }}
{{3!}} + \tfrac{{u^3 }}
{{4!}} \mp ...}}
{{e^u - 1}} \cdot du}
$

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

So by Lemma 3of the previous post: $
\int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
{{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \tfrac{1}
{2} \cdot \zeta \left( 2 \right) - \tfrac{1}
{3} \cdot \zeta \left( 3 \right) \pm ...
$
And finally, by the Lemma we've just seen: $
\int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
{{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \gamma \blacksquare
$