Hot Integral

Our Integral can be written as:

$\displaystyle I= \int_0^{\infty} \frac{x^{n-1}}{x-1}dx = \int_0^{1^-} \frac{x^{n-1}}{x-1}dx + \int_{1^+}^{\infty} \frac{x^{n-1}}{x-1}dx$

Use the substitution $x\to \frac{1}{x}$ in the second integral, to get:

$\displaystyle I = - \int_0^{1^-} \frac{x^{n-1}}{1-x}dx + \int_0^{1^-} \frac{x^{-n}}{1-x}dx$

Now as long as the integral limits are the same, and bounded between $0$ and $1$ , we can rewrite $I$ as:

$\displaystyle I= - \int_0^{1^-} \sum_{m=0}^{\infty} \bigg( x^{m+n-1} - x^{m-n} \bigg) dx$

Integrating:

$\displaystyle I = - \sum_{m=0}^{\infty} \bigg( \frac{1}{m+n} + \frac{1}{m-n+1} \bigg)$

$\displaystyle = \sum_{m=1}^{\infty} \frac{1}{m+(1-n)} - \sum_{m=1}^{\infty} \frac{1}{m+n} - \frac{1}{n} +\frac{1}{1-n}$

Now we are going to show that :

$\displaystyle \sum_{m=1}^{\infty} \frac{1}{m+(1-n)} - \sum_{m=1}^{\infty} \frac{1}{m+n} = H_n - H_{1-n}$

Where $H_k$ is the $k^{th}$ Harmonic Number, Generalized to the positive non-integral arguments.

Recall that for any real positive $x$ :

$\displaystyle H_x = x\sum_{k=1}^{\infty} \frac{1}{k(k+x)}$

$\displaystyle = \sum_{k=1}^{\infty} \frac{1}{k} - \sum_{k=1}^{\infty} \frac{1}{k+x}$

Therefore:

$\displaystyle H_n - H_{1-n} = [\sum_{k=1}^{\infty} \frac{1}{k} - \sum_{k=1}^{\infty} \frac{1}{k+n} ] - [\sum_{k=1}^{\infty} \frac{1}{k} - \sum_{k=1}^{\infty} \frac{1}{k+(1-n)} ]$

$\displaystyle = \sum_{k=1}^{\infty} \frac{1}{k+(1-n)} - \sum_{k=1}^{\infty} \frac{1}{k+n}$

So rewrite $I$ :

$\displaystyle I =H_n - H_{1-n} - \frac{1}{n} +\frac{1}{1-n}$

$\bullet \mathbf{ \; Preposition \; 1 \; : }$

For $0<r<1$ , we have the following Reflection Relation:

$\displaystyle H_{1-r} - H_r = \pi \cot (\pi r ) -\frac{1}{r} +\frac{1}{1-r}$

To prove this, we need other prepositions:

$\bullet \mathbf{ \; Preposition \; 2 \; : }$

For any positive real $y$ :

$\displaystyle \int_0^y H_x dx = y\gamma + \ln (y!)$

Where $\gamma$ is the Euler-Mascheroni Constant .

$\mathbf{ \; Proof \; : }$

Start with the definition of the harmonic number, and integrate both sides :

$\displaystyle \int_0^y H_x dx = \int_0^y x\sum_{k=1}^{\infty} \frac{1}{k(k+x)}dx$

(just basic integration and evaluation):

$\displaystyle = \lim_{i\to \infty} \sum_{k=1}^i \bigg( \frac{y}{k} - \ln (y+k) +\ln k \bigg)$

$\displaystyle = \lim_{i\to \infty} ( yH_i - \ln (\frac{(y+i)!}{y!}) + \ln i! )$

$\displaystyle = \lim_{i\to \infty} ( yH_i + (- y\ln i + y\ln i) - \ln (\frac{(y+i)!}{y!}) + \ln i! )$

$\displaystyle = \lim_{i\to \infty} ( y{\color{#20A900}{(H_i - \ln i )}} + {\color{#D61F06}{\ln \Bigg(\frac{i^y i!}{(y+i)!}\Bigg)}} + \ln y! )$

Now, Recall the Definition of the Euler-Mascheroni Constant. Then notice that the red term will vanish using Stirling's Formula. So:

$\displaystyle \boxed{ \int_0^y H_x dx = y\gamma + \ln y! }$

$\bullet \mathbf{ \; Preposition \; 3 \; : }$

For any $n>0$ :

$\displaystyle \psi (n) = H_{n} - \frac{1}{n} - \gamma$

Where : $\psi (n)$ is the Digamma Function , and $\gamma$ is the Euler-Mascheroni Constant .

$\mathbf{ \; Proof \; : }$

Using the identity #2 :

$\displaystyle \int_0^y H_x dx = y\gamma + \ln \Gamma (y+1)$

Differentiate w.r.t $y$ Then Recall the definition of the Digamma function :

$\displaystyle H_y = \gamma + \psi (y+1)$

Therefore : ( $y \to (n-1)$ )

$\displaystyle \boxed{\psi (n) = H_{n-1} - \gamma = H_{n} - \frac{1}{n} - \gamma}$

Now, we can prove our first preposition.

Recall the Well Known Euler's Reflection Formula for the Gamma Function:

$\displaystyle \Gamma (z) \Gamma (1-z) = \pi \csc (\pi z)$

Take the Natural Logarithm of both Sides:

$\displaystyle \ln \Gamma (z) + \ln \Gamma (1-z) = \ln \pi - \ln \sin (\pi z)$

Differentiate Both Sides:

$\displaystyle \psi (z) - \psi (1-z) = -\pi \cot (\pi z)$

Now use the Identity #3 :

$\displaystyle H_z -\frac{1}{z} - H_{1-z} + \frac{1}{1-z} = -\pi \cot (\pi z)$

Rearranging, the reflection relation is thus proved:

$\displaystyle \boxed{H_{1-r} - H_r = \pi \cot (\pi r ) -\frac{1}{r} +\frac{1}{1-r} }$

Now evaluate this relation back in the equation of $I$ :

$\displaystyle \boxed{I = -\frac{\pi}{\tan (\pi n ) } }$

The rest work is pretty easy, use $\tan \frac{\pi}{8} = \sqrt{2} -1$ , and the final answer becomes: $\boxed{ \pi + \ln (\sqrt{2} -1 ) }$ .

Nice problem, seriously.

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

is our problem. Well, I will do it using contour integration.

What we can see? A clear pole at $\displaystyle z = 1$ (we will just change all the x to z since we are in Argand plane now) and there is a branch point as well $\displaystyle z = 0$ . That were the singularities and now we have to play with them.

Consider the contour as a full circle around the origin with a branch cut towards the positive real axis, in short, a "keyhole contour".

So, $\displaystyle C = \Gamma + {L}_{1} + {L}_{2} + \gamma$

$\Gamma$ represents the "outside" of C with radius M( $\displaystyle M \Rightarrow \infty$ )

$\displaystyle \gamma$ represents the small circle around the origin with radius $\displaystyle \epsilon(\Rightarrow 0)$

$\displaystyle {L}^{1}$ represents the straight line extending from the small circle over the +ve real axis.

$\displaystyle {L}_{2}$ represents the straight line extending from the small circle below the +ve real axis.

We can see that

$\displaystyle \int_{\Gamma}{\frac{{z}^{n-1}}{z-1}} \Rightarrow 0$

$\displaystyle \int_{\gamma}{\frac{{z}^{n-1}}{z-1}} \Rightarrow 0$

*I am not giving the details of the above integrals but they can easily be seen.

And surely, we'll then evaluate the rest.

We can write ${z}^{n-1} = {e}^{(n-1)ln(z)}$ . Why did we do that?

$\displaystyle \int_{{L}_{1}}{\frac{{z}^{n-1}}{z-1}} = \int_{0}^{\infty}{\frac{{e}^{(n-1)ln(z)}}{z-1}}$

as $\displaystyle {L}_{1}$ is a straight line.

Similarly,

$\displaystyle \int_{{L}_{1}}{\frac{{z}^{n-1}}{z-1}} = -\int_{0}^{\infty}{\frac{{e}^{(n-1)(ln(z({e}^{2i\pi}))}}{z-1}}$

We did that for this. Now, it becomes easy for us to get the answer.

Why did the minus sign come in front? $\displaystyle {L}_{2}$ is below so, the denominator would change as well.

Why did that $\displaystyle {e}^{2i\pi}$ come? Well, because that is the "phase difference" between the two.

Finally, we know that

$\displaystyle C = \Gamma + {L}_{1} + {L}_{2} + \gamma$

Hence,

$\displaystyle \int_{C}{\frac{{z}^{n-1}}{z-1}} = (1 - {e}^{2(n-1)i\pi})(\int_{0}^{\infty}{\frac{{z}^{n-1}}{z-1}})$

Now, the line integral around the curve C can be found using Residue theorem and comes out to be $\displaystyle 2i\pi{e}^{2i\pi(n-1)}$

Therefore,

$\displaystyle \int_{0}^{\infty}{\frac{{z}^{n-1}}{z-1}} = \frac{2i\pi{e}^{2i\pi(n-1)}}{(1 - {e}^{2(n-1)i\pi})}$

which can be simplified to

$\displaystyle = -\frac{\pi}{tan(n\pi)} - i\pi$

Now, I don't how to get rid of that extra $i\pi$ and it was the reason of I first reporting it. But still, we can see that we want a real value and consider the first. Help me here!