1. ## cool integral

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

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

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

But that is certainly not the solution.

It has the same solution as:

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

2. 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.

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

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

But that is certainly not the solution.

It has the same solution as:

$\displaystyle \int_{0}^{1}\frac{1-e^{-x}-e^{\frac{-1}{x}}}{x}dx$
I have worked on this integral for about two weeks and have finally come up with a solution

First split it into

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

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

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

Integration by parts on both gives

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

Now consider the gamma function

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

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

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

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

The value of Gamma' evaluated at one...I cannot prove..haha

Here is a tidbit you may find nice:

The DiGamma function:

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

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

The DiGamma function:

$\displaystyle {\Psi}(1)=\frac{{\Gamma}(1)'}{{\Gamma}(1)}=-1-{\gamma}+\sum_{r=1}^{\infty}\left(\frac{1}{r}-\frac{1}{r+1}\right)=-{\gamma}$
Thanks Galctus...I have heard of the Digamma function but I was unaware of its series representation...I just knew that $\displaystyle \Gamma'(1)=-\gamma$

5. Lemma 1. Let $\displaystyle f:\mathbb{R}/\mathbb{R}^ - \to \mathbb{R}$ be a function with continuous derivative in its domain.

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

Proof

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

So: $\displaystyle \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)} $$\displaystyle = \sum\limits_{k = 0}^{n - 1} {\left( {k + 1 - 1} \right) \cdot f\left( {k + 1} \right)} - \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( k \right)}$$\displaystyle = \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)} $$\displaystyle = n \cdot f\left( n \right) - \sum\limits_{k = 1}^n {f\left( k \right)} Now integrating by parts: \displaystyle \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.\displaystyle \square Lemma 2. We have: \displaystyle \mathop {\lim }\limits_{s \to 1^ + } \left( {\zeta \left( s \right) - \tfrac{1} {{s - 1}}} \right) = \gamma Proof First take \displaystyle f\left( x \right) = \tfrac{1} {{x + 1}} in Lemma 1: We get \displaystyle \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: \displaystyle \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 \displaystyle n \to + \infty - this in fact shows the existance of the gamma constant- then by definition of the gamma constant: \displaystyle \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 \displaystyle f\left( x \right) = \tfrac{1} {{\left( {x + 1} \right)^s }} with \displaystyle s>1 in Lemma 1 : \displaystyle \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: \displaystyle \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 \displaystyle \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: \displaystyle \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 \displaystyle \square Lemma 3. \displaystyle \zeta \left( s \right) \cdot \Gamma \left( s \right) = \int_0^\infty {\tfrac{{x^{s - 1} }} {{e^x - 1}}dx} : \displaystyle s>1 Proof: \displaystyle \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 \displaystyle \int_0^\infty {x^{s - 1} \cdot e^{ - nx} dx} = \tfrac{{\Gamma \left( s \right)}} {{n^s }} we are done \displaystyle \square Proof : \displaystyle \int_0^1 {\left( {\tfrac{1} {{1 - x}} + \tfrac{1} {{\log \left( x \right)}}} \right)dx} = \gamma Let \displaystyle I be the integral then letting \displaystyle u = - \log \left( x \right) we get \displaystyle I = \int_0^\infty {e^{ - u} \cdot \left( {\tfrac{1} {{1 - e^{ - u} }} - \tfrac{1} {u}} \right)du} Now note that: \displaystyle \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}} ( \displaystyle s>1 ) Thus: \displaystyle \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: \displaystyle \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: \displaystyle \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$$\displaystyle = \int_0^\infty {e^{ - u} \cdot \left( {\tfrac{1} {{1 - e^{ - u} }} - \tfrac{{1 }} {u}} \right)} du \square$

6. Lemma. $\displaystyle \sum\limits_{k = 2}^\infty {\tfrac{{\zeta \left( k \right)}} {k} \cdot \left( { - 1} \right)^k } = \gamma$

Proof

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

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

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

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

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

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

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

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

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