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Laplace transform of a function!

Calculus (Integrals, Series)
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Tolaso J Kos
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Laplace transform of a function!

#1

Post by Tolaso J Kos » Sun Apr 24, 2016 6:42 pm

Prove that:

$$\int_{0}^{\infty}e^{-x} \ln \left ( \ln \left ( e^x+ \sqrt{e^{2x}-1} \right ) \right )\, {\rm d}x = -\gamma+ 4 \log \Gamma \left ( \frac{1}{4} \right )-3\log 2 -2 \log \pi$$

where $\gamma$ stands for the Euler - Mascheroni constant and $\Gamma$ is the Gamma function.
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Re: Laplace transform of a function!

#2

Post by whitexlotus » Sun Sep 04, 2016 4:08 pm

Tolaso J Kos wrote:Prove that:

$$\int_{0}^{\infty}e^{-x} \ln \left ( \ln \left ( e^x+ \sqrt{e^{2x}-1} \right ) \right )\, {\rm d}x = -\gamma+ 4 \log \Gamma \left ( \frac{1}{4} \right )-3\log 2 -2 \log \pi$$

where $\gamma$ stands for the Euler - Mascheroni constant and $\Gamma$ is the Gamma function.
Hi tolaso, I'm pprime (http://artofproblemsolving.com/community/c89" onclick="window.open(this.href);return false;), I do not have much free time, so write briefly my solution

$$\int\limits_{0}^{+\infty }{e^{-x}\ln \ln \left( e^{x}+\sqrt{e^{2x}-1} \right)dx}=\int\limits_{0}^{+\infty }{e^{-x}\ln \left( \cosh ^{-1}e^{x} \right)dx}=\int\limits_{1}^{+\infty }{\frac{\ln \left( \cosh ^{-1}e^{x} \right)}{x^{2}}dx}$$

$$=\int\limits_{0}^{+\infty }{\frac{\sinh x\ln x}{\cosh ^{2}x}dx}$$

Consider
$$F\left( s \right)=\int\limits_{0}^{+\infty }{\frac{x^{s}\sinh x}{\cosh ^{2}x}dx}$$ and note that $$F'\left( 0 \right)=\int\limits_{0}^{+\infty }{\frac{\sinh x\ln x}{\cosh ^{2}x}dx}=\int\limits_{0}^{+\infty }{e^{-x}\ln \ln \left( e^{x}+\sqrt{e^{2x}-1} \right)dx}$$

then
$$F\left( s \right)=\int\limits_{0}^{+\infty }{\frac{x^{s}\sinh x}{\cosh ^{2}x}dx}=s\int\limits_{0}^{+\infty }{\frac{x^{s-1}}{\cosh x}dx}=2s\int\limits_{0}^{+\infty }{\frac{e^{-x}x^{s-1}}{1+e^{-2x}}dx}$$

$$=2s\int\limits_{0}^{+\infty }{e^{-x}x^{s-1}\sum\limits_{n=0}^{+\infty }{\left( -e^{-2x} \right)^{n}}dx=}2s\sum\limits_{n=0}^{+\infty }{\left( -1 \right)^{n}\int\limits_{0}^{+\infty }{e^{-x-2nx}x^{s-1}dx}}$$

$$=2s\Gamma \left( s \right)\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( 1+2n \right)^{s}}}=2^{1-s}s\Gamma \left( s \right)\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( \frac{1}{2}+n \right)^{s}}}$$

$$S=\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( \frac{1}{2}+n \right)^{s}}}=\frac{\left( -1 \right)^{0}}{\left( \frac{1}{2}+0 \right)^{s}}+\frac{\left( -1 \right)^{1}}{\left( \frac{1}{2}+1 \right)^{s}}+\frac{\left( -1 \right)^{2}}{\left( \frac{1}{2}+2 \right)^{s}}+\frac{\left( -1 \right)^{3}}{\left( \frac{1}{2}+3 \right)^{s}}+\frac{\left( -1 \right)^{4}}{\left( \frac{1}{2}+4 \right)^{s}}+\frac{\left( -1 \right)^{5}}{\left( \frac{1}{2}+5 \right)^{s}}+...$$

$$=\left( \frac{1}{\left( \frac{1}{2} \right)^{s}}+\frac{1}{\left( \frac{1}{2}+2 \right)^{s}}+\frac{1}{\left( \frac{1}{2}+4 \right)^{s}}+... \right)-\left( \frac{1}{\left( \frac{1}{2}+1 \right)^{s}}+\frac{1}{\left( \frac{1}{2}+3 \right)^{s}}+\frac{1}{\left( \frac{1}{2}+5 \right)^{s}}+... \right)$$

$$=\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( \frac{4n+1}{2} \right)^{s}}}-\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( \frac{4n+3}{2} \right)^{s}}}=\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( 2n+\frac{1}{2} \right)^{s}}}-\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( 2n+\frac{3}{2} \right)^{s}}}$$

$$=\frac{1}{2^{s}}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{1}{4} \right)^{s}}}-\frac{1}{2^{s}}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{3}{4} \right)^{s}}}=\frac{1}{2^{s}}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$F\left( s \right)=2^{1-s}s\Gamma \left( s \right)\sum\limits_{n=0}^{+\infty }{\frac{\left( -1 \right)^{n}}{\left( \frac{1}{2}+n \right)^{s}}}=2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$=2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$F'\left( s \right)=\frac{d}{ds}2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$=\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\frac{d}{ds}\left( 2^{1-2s}\Gamma \left( s+1 \right) \right)+2^{1-2s}\Gamma \left( s+1 \right)\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

$$=2^{1-2s}\Gamma \left( s+1 \right)\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\left( \psi ^{\left( 0 \right)}\left( s+1 \right)-2\log 2 \right)+2^{1-2s}\Gamma \left( s+1 \right)\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)$$

where $$\zeta \left( s,q \right)=\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( q+n \right)^{s}}}$$
$$\frac{d}{ds}\zeta \left( s,\frac{1}{4} \right)=\frac{d}{ds}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{1}{4} \right)^{s}}}=-\left( \frac{\log \left( \frac{1}{4} \right)}{\left( \frac{1}{4} \right)^{s}}+\frac{\log \left( 1+\frac{1}{4} \right)}{\left( 1+\frac{1}{4} \right)^{s}}+\frac{\log \left( 2+\frac{1}{4} \right)}{\left( 2+\frac{1}{4} \right)^{s}}+\frac{\log \left( 3+\frac{1}{4} \right)}{\left( 3+\frac{1}{4} \right)^{s}}+... \right)$$

$$\frac{d}{ds}\zeta \left( s,\frac{3}{4} \right)=\frac{d}{ds}\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( n+\frac{3}{4} \right)^{s}}}=-\left( \frac{\log \left( \frac{3}{4} \right)}{\left( \frac{3}{4} \right)^{s}}+\frac{\log \left( 1+\frac{3}{4} \right)}{\left( 1+\frac{3}{4} \right)^{s}}+\frac{\log \left( 2+\frac{3}{4} \right)}{\left( 2+\frac{3}{4} \right)^{s}}+\frac{\log \left( 3+\frac{3}{4} \right)}{\left( 3+\frac{3}{4} \right)^{s}}+... \right)$$

$$F'\left( 0 \right)=2\Gamma \left( 1 \right)\left( \zeta \left( 0,\frac{1}{4} \right)-\zeta \left( 0,\frac{3}{4} \right) \right)\left( \psi ^{\left( 0 \right)}\left( 1 \right)-2\log 2 \right)+2\Gamma \left( 1 \right)\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\left| _{s=0} \right.$$

$$F'\left( 0 \right)=-\left( \gamma +2\log 2 \right)+2\frac{d}{ds}\left( \zeta \left( s,\frac{1}{4} \right)-\zeta \left( s,\frac{3}{4} \right) \right)\left| _{s=0} \right.$$

$$=-\left( \gamma +2\log 2 \right)-2\log \left( \frac{\Gamma \left( \frac{3}{4} \right)}{\Gamma \left( \frac{1}{4} \right)} \right)=-\left( \gamma +2\log 2 \right)-2\log \left( \frac{\sqrt{2}\pi }{\Gamma ^{2}\left( \frac{1}{4} \right)} \right)$$

$$=-\gamma -3\log 2-2\log \pi +4\log \Gamma \left( \frac{1}{4} \right)$$

Done (:
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Re: Laplace transform of a function!

#3

Post by Tolaso J Kos » Mon Sep 05, 2016 8:30 am

Hi pprime. Thank you for accepting my invitation and joined the forum. Hope you enjoy it!
Regards,

T.
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