Fourier series of Loggamma function - Part 2

         

         This post is part 2 of this post, where we started to derive the fourier series of the Loggamma function. Today we are going to evaluate the coefficient b_{k}  and then, put all the pieces together to find the final expression.

 Evaluation of  b_{k}

Consider

I=\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)dx

Integrating by parts

u= \log(\Gamma(x)) \, \Rightarrow du= \psi(x) dx

dv=\sin(2 \pi k x)dx \, \Rightarrow v= - \frac{\cos(2 \pi k x)}{2 \pi k}

\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)dx=- \frac{\cos(2 \pi k x)}{2 \pi k}\log(\Gamma(x)) \Big|_{0}^{1}+\frac{1}{2 \pi k}\int_{0}^{1}\psi(x)\cos(2 \pi k x)dx

now observe the following with regards to the limit:

- \frac{\cos(2 \pi k x)}{2 \pi k}\log(\Gamma(x)) \Big|_{0}^{1}=-\lim_{x \rightarrow 1} \cdot\frac{\cos(2 \pi k x)}{2 \pi k} \cdot \lim_{x \rightarrow 1} \cdot\log(\Gamma(x))+\lim_{x \rightarrow 0} \cdot\frac{\cos(2 \pi k x)}{2 \pi k} \cdot \lim_{x \rightarrow 0} \cdot\log(\Gamma(x))

=-\frac{1}{2 \pi k} \cdot \lim_{x \rightarrow 1} \cdot\log(\Gamma(x))+\frac{1}{2 \pi k} \cdot \lim_{x \rightarrow 0} \cdot\log(\Gamma(x))

-\frac{1}{2 \pi k}\int_{0}^{1}\psi(x)dx

therefore


\boxed{\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)dx=-\frac{1}{2 \pi k}\int_{0}^{1}\psi(x)(1-\cos(2 \pi k x))dx}(1)

Recall now the integral representation of the Digamma function


\boxed{\psi(x)=\int_{0}^{\infty}\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}dt}(2)


Plugging (2) on the right hand side of (1)

I=-\frac{1}{2 \pi k}\int_{0}^{1}(1-\cos(2 \pi k x)) \, \bigg\{\int_{0}^{\infty}\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}dt \bigg\}\,dx


I=-\frac{1}{2 \pi k}\int_{0}^{\infty} \!\!\!\color{blue}{\int_{0}^{1}(1-\cos(2 \pi k x)) \, \bigg[\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}} \bigg]\,dx}\color{black} \, dt(3)


Lets now concentrate in the inner integral in blue and break it down in two integrals, namely


J_{1}=\int_{0}^{1} \frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}} \,dx


J_{2}=-\int_{0}^{1}\cos(2 \pi k x) \, \bigg[\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}} \bigg]\,dx


Lest begin with

J_{1}=\int_{0}^{1} \frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}} \,dx=\int_{0}^{1} \frac{e^{-t}}{t}dx-\int_{0}^{1}\frac{e^{-xt}}{1-e^{-t}} \,dx

=\frac{e^{-t}}{t}-\frac{1}{1-e^{-t}}\int_{0}^{1} e^{-xt}\,dx

=\frac{e^{-t}}{t}-\frac{1}{1-e^{-t}} \bigg[ \,\,\,\frac{e^{-xt}}{t}\Big|_{0}^{1} \,\,\,\, \bigg]

=\frac{e^{-t}}{t}-\frac{1}{(1-e^{-t})} \frac{(1-e^{-t})}{t}

\boxed{J_{1}=\frac{e^{-t}}{t}-\frac{1}{t}}(4)

Now

J_{2}=-\int_{0}^{1}\cos(2 \pi k x) \, \bigg[\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}} \bigg]\,dx

=-\frac{e^{-t}}{t}\int_{0}^{1}\cos(2 \pi k x) \,dx+\frac{1}{1-e^{-t}} \int_{0}^{1}\cos(2 \pi k x)e^{-xt}\,dx

=-\frac{e^{-t}}{t}\underbrace{\Big[ \frac{\sin(2 \pi k x)}{2 \pi k}\Big|_{0}^{1} \,\,\,\Big]}_{=0}+\frac{1}{1-e^{-t}} \int_{0}^{1}\cos(2 \pi k x)e^{-xt}\,dx

=\frac{1}{1-e^{-t}} \int_{0}^{1}\cos(2 \pi k x)e^{-xt}\,dx

This integral was already evaluated here, therefore

J_{2}=\frac{1}{(1-e^{-t})}\frac{t \cdot (1-e^{-t})}{t^2 + (2 \pi k )^2}

\boxed{J_{2}=\frac{t }{t^2 + (2 \pi k )^2}}(5)

Plugging (4) and (5) back in (3), we get

I=-\frac{1}{2 \pi k}\int_{0}^{\infty}\bigg[\frac{e^{-t}}{t}-\frac{1}{t}+ \frac{t }{t^2 + (2 \pi k )^2} \bigg]dt

=-\frac{1}{2 \pi k}\bigg[\int_{0}^{\infty}\frac{e^{-t}}{t}dt-\int_{0}^{\infty}\frac{1}{t}dt+ \int_{0}^{\infty}\frac{t }{t^2 + (2 \pi k )^2} dt\bigg](6)

We will consider each of these integrals separately


K_{1}=\int_{0}^{\infty}\frac{e^{-t}}{t}dt

K_{2}=\int_{0}^{\infty}\frac{1}{t}dt

K_{3}=\int_{0}^{\infty}\frac{t }{t^2 + (2 \pi k )^2} dt


To compute  K_{1}  we perform an integration by parts

K_{1}=\int_{0}^{\infty}\frac{e^{-t}}{t}dt= (u.v)\Big|_{0}^{\infty}-\int_{0}^{\infty}vdu

u=e^{-t} \Rightarrow \, du= - e^{-t}dt

dv=\frac{dt}{t} \, \Rightarrow \, v= \log(t)

K_{1}=(e^{-t}\log(t))\Big|_{0}^{\infty}+\int_{0}^{\infty}e^{-t}\log(t)

\boxed{K_{1}=(e^{-t}\log(t))\Big|_{0}^{\infty}-\gamma}(7)


Where in the last line we used the result    \int_{0}^{\infty}e^{-t}\log(t)= -\gamma   , that will be proved in a future post.

\gamma is the Euler-Mascheroni constant.

\boxed{K_{2}=\int_{0}^{\infty}\frac{1}{t}dt= \log(t)\Big|_{0}^{\infty}}(8)

K_{3}=\int_{0}^{\infty}\frac{t }{t^2 + (2 \pi k )^2} dt

To evaluate K_{3}, we should consider the following indefinite integral first:

\int\frac{x}{a^2+x^2}dx

substituting u=a^2+x^2 \, \Rightarrow \, du=2dx \,\, \Rightarrow dx= \frac{du}{2}

\int\frac{x}{a^2+x^2}dx=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\log(u)+C=\frac{1}{2}\log(a^2+x^2)+C

We conclude therefore that

\boxed{K_{3}=\int_{0}^{\infty}\frac{t }{t^2 + (2 \pi k )^2} dt=\frac{1}{2}\log(t^2+(2 \pi k)^2\Big|_{0}^{\infty}}(9)

Putting (7),(8) and (9) in (6) we get

I=-\frac{1}{2 \pi k}\bigg[ \,\, \Big(e^{-t}\log(t)\Big|_{0}^{\infty}-\gamma\Big)-\Big(\log(t)\Big|_{0}^{\infty}\Big)+\frac{1}{2}\Big(\log(t^2+(2 \pi k)^2)\Big|_{0}^{\infty}\Big)\,\,\,\,\bigg]


=\frac{\gamma}{2 \pi k}-\frac{1}{2 \pi k}\lim_{t \rightarrow \infty}\bigg[ \,\, e^{-t}\log(t)-\log(t)+\frac{1}{2}\log(t^2+(2 \pi k)^2)\,\,\,\,\bigg]+ \\ +\frac{1}{2 \pi k}\lim_{t \rightarrow 0}\bigg[ \,\, e^{-t}\log(t)-\log(t)+\frac{1}{2}\log(t^2+(2 \pi k)^2)\,\bigg]


=\frac{\gamma}{2 \pi k}-\frac{1}{2 \pi k}\bigg[ \,\,\underbrace{\lim_{t \rightarrow \infty} e^{-t}\log(t)}_{=0}-\lim_{t \rightarrow \infty}\log(t)+\frac{1}{2}\lim_{t \rightarrow \infty}\log((t^2)(1+\frac{(2 \pi k)^2}{t^2})\,\,\,\,\bigg]+ \\ +\frac{1}{2 \pi k}\bigg[ \,\,\underbrace{\lim_{t \rightarrow 0} e^{-t}\log(t)}_{=\log(t)}-\lim_{t \rightarrow 0}\log(t)+\frac{1}{2}\lim_{t \rightarrow 0}\log(t^2+(2 \pi k)^2)\,\bigg]


=\frac{\gamma}{2 \pi k}-\frac{1}{2 \pi k}\bigg[ \,\,-\lim_{t \rightarrow \infty}\log(t)+\frac{1}{2}\lim_{t \rightarrow \infty}\log((t^2)+\frac{1}{2}\underbrace{\lim_{t \rightarrow \infty}\log(1+\frac{(2 \pi k)^2}{t^2})}_{=0}\,\,\,\,\bigg]+ \\ +\frac{1}{2 \pi k}\bigg[ \,\,\frac{1}{2}\log((2 \pi k)^2)\,\bigg]


=\frac{\gamma}{2 \pi k}-\frac{1}{2 \pi k}\bigg[ \,\,-\lim_{t \rightarrow \infty}\log(t)+\lim_{t \rightarrow \infty}\log(t)\,\,\bigg] +\frac{\log(2 \pi k)}{2 \pi k}

And Finally!!


\boxed{\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)dx=\frac{\gamma+\log(2 \pi k)}{2 \pi k}}(10)

To find b_{k} we should substitue (10) in the expression below

b_{k}= 2\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)dx


\boxed{b_{k}=\frac{\gamma+\log(2 \pi k)}{ \pi k}}}

Lets put now all the pieces together. We have:

a_{0}= \log(2 \pi)

a_{k}=\frac{1}{2k}

b_{k}=\frac{\gamma+\log(2 \pi k)}{ \pi k}}

\log(\Gamma(x))=\frac{a_{0}}{2}+ \sum_{k=1}^{\infty}a_{k}\cos(2 \pi k x) + \sum_{k=1}^{\infty}b_{k}\sin(2 \pi k x)



\boxed{\log(\Gamma(x))=\frac{\log(2 \pi)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos(2 \pi k x) + \sum_{k=1}^{\infty}\frac{\gamma+\log(2 \pi k)}{ \pi k}\sin(2 \pi k x)}(11)


We can rearrange (11) to put it in a more familiar way

\log(\Gamma(x))=\frac{\log(2 \pi)}{2}+ \frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k}\cos(2 \pi k x) + \frac{\gamma}{\pi}\sum_{k=1}^{\infty}\frac{\sin(2 \pi k x)}{ k}+ \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log(2 \pi k)\sin(2 \pi k x)}{ k}(12)

From this post we know that

\sum_{k=1}^{\infty}\frac{\sin(2 \pi k x)}{k} =\pi\Big\{\frac{1}{2}- x\Big\}(13)

\sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k} = -\log(2 \sin( \pi x))(14)

substituting (13) and (14) in (12)


\log(\Gamma(x))=\frac{\log(2 \pi)}{2}- \frac{1}{2}\log(2 \sin( \pi x)) + \gamma\Big\{\frac{1}{2}- x\Big\}+ \frac{\log( 2 \pi)}{\pi}\sum_{k=1}^{\infty}\frac{\sin(2 \pi k x)}{ k}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log( k)\sin(2 \pi k x)}{ k}


\log(\Gamma(x))=\frac{\log(2 \pi)}{2}- \frac{1}{2}\log(2 \sin( \pi x)) + \gamma\Big\{\frac{1}{2}- x\Big\}+ \log( 2 \pi)\Big\{\frac{1}{2}- x\Big\}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log( k)\sin(2 \pi k x)}{ k}


\log(\Gamma(x))=\frac{\log(2 )}{2}+\frac{\log(\pi)}{2}-\frac{\log(2)}{2}-\frac{\log(\sin(\pi x))}{2}+ \gamma\Big\{\frac{1}{2}- x\Big\}+\log(2)\Big\{\frac{1}{2}- x\Big\}+\log(\pi)\Big\{\frac{1}{2}- x\Big\}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log( k)\sin(2 \pi k x)}{ k}


And we arrive to the more familiar expression


\boxed{\log(\Gamma(x))= \Big(\gamma+\log(2)\Big)\Big( \frac{1}{2}- x\Big)+(1-x)\log(\pi)-\frac{\log(\sin(\pi x))}{2}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log( k)\sin(2 \pi k x)}{ k}}


Ricardo Albahari

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