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[SOLVED] The Laplace of unfamiliar function

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I found the following integral


\(\displaystyle \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt \)

This thread will be dedicated to solve the integral , any ideas will always be welcomed (Cool).

Serious steps will be made in the next threads . It seems to involve Laplace identities the value returned by Mathematica is insane (Sweating).
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt \)

We can start by the following integral

\(\displaystyle

\begin{align*}

I(s) = \int_{0}^{\infty} t^{s-1} \, e^{-at} \, \sin(bt) dt

&= \int_{0}^{\infty}t^{s-1} \, e^{-at}\sum_{n\geq 0} \frac{(-1)^n (bt)^{2n+1}}{\Gamma(2n+2)} \\

&= \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1}}{\Gamma(2n+2)} \int_{0}^{\infty}t^{s+2n} \, e^{-at} dt \\

&= \frac{1}{a^s} \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1} \Gamma (s+2n+1)}{\Gamma(2n+2) a^{2n+1}}

\end {align*} \)


\(\displaystyle

\begin{align*}

I'(0) = \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt

&= \sum_{n\geq 0} \frac{(-1)^n \psi_0(2n+1)}{2n+1} \left( \frac{b}{a} \right)^{2n+1} -\log(a) \sum_{n\geq 0} \frac{(-1)^n }{2n+1} \left( \frac{b}{a} \right)^{2n+1}\\

&= \sum_{n\geq 0} \frac{(-1)^n H_{2n}-\gamma}{2n+1} \left( \frac{b}{a} \right)^{2n+1} -\log(a) \arctan \left( \frac{b}{a} \right) \\

&= \sum_{n\geq 0} \frac{(-1)^n H_{2n}}{2n+1} \left( \frac{b}{a} \right)^{2n+1}-(\gamma +\log(a))\arctan \left( \frac{b}{a} \right)

\end{align*}
\)

It remains to solve the following

\(\displaystyle \sum_{n\geq 0} \frac{(-1)^n H_{2n} }{2n+1} \, x^{2n+1}\)

I will try to evaluate the Harmonic sum in the next post ,,,
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
After lots of thinking here is a solution to the harmonic sum

\(\displaystyle
\begin{align*}

\sum_{k\geq 0}(-1)^k H_{2k} x^{2k}&= \sum_{k\geq 0}(-1)^k x^{2k} \int^1_0 \frac{1-t^{2k}}{1-t} \, dt\\

&= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k x^{2k} \left(1-t^{2k}\right) \, dt\\
&= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k \left(x^{2k}-(xt)^{2k}\right) \, dt\\
&= \int^1_0 \frac{1}{1-t}\left(\frac{1}{1+x^2}-\frac{1}{1+t^2x^2}\right) \, dt\\
&=\frac{1}{1+x^2} \int^1_0 \frac{1+t^2x^2-1-x^2}{(1-t)(1+t^2x^2)} \, dt\\
&=\frac{-x^2}{1+x^2} \int^1_0 \frac{(1-t^2)}{(1-t)(1+t^2x^2)} \, dt\\
&=\frac{-x^2}{1+x^2} \int^1_0 \frac{1+t}{(1+t^2x^2)} \, dt\\
&=\frac{-x^2}{1+x^2} \left( \int^1_0 \frac{1}{1+t^2x^2}+\frac{t}{1+t^2x^2} \, dt \right) \\
&= \frac{-1}{2(1+x^2)} \left(2x \arctan (x) + \log(1+x^2) \right) \\

\end{align*}
\)

Using this we conclude by integrating

\(\displaystyle \sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} x^{2k}=-\frac{1}{2} \log(1+x^2) \arctan(x) \)

Hence the following

\(\displaystyle \sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} \left(\frac{b}{a} \right)^{2k+1}=-\frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) \arctan \left(\frac{b}{a} \right) \)

\(\displaystyle
\begin{align*}

\int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt &= -\left( \frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) + \gamma +\log(a) \right) \arctan \left( \frac{b}{a} \right)\\
&=- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b}{a} \right)\\

\end{align*}
\)

Which is the result we are seeking for .... (Happy)
 
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