Question Number 150259 by mathdanisur last updated on 10/Aug/21
$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{e}^{−\boldsymbol{\mathrm{st}}} \left(\mathrm{cosh}\left(\mathrm{2t}\right)−\mathrm{cosh}\left(\mathrm{5t}\right)\right)\mathrm{dt}}{\mathrm{t}}=? \\ $$
Answered by Ar Brandon last updated on 10/Aug/21
![I(s)=∫_0 ^∞ ((e^(−st) (cosh(2t)−cosh(5t)))/t)dt I ′(s)=−∫_0 ^∞ e^(−st) (cosh(2t)−cosh(5t))dt =−(1/2)∫_0 ^∞ e^(−st) (e^(2t) +e^(−2t) −e^(5t) −e^(−5t) )dt =−(1/2)[(e^(−(s−2)t) /(2−s))−(e^(−(s+2)t) /(s+2))−(e^(−(s−5)t) /(5−s))+(e^(−(s+5)t) /(s+5))]_0 ^∞ =(1/2)((1/(2−s))−(1/(s+2))−(1/(5−s))+(1/(s+5))) I(s)=(1/2)(ln∣s+5∣+ln∣s−5∣−ln∣s+2∣−ln∣s−2∣)+C =(1/2)ln∣((s^2 −25)/(s^2 −4))∣+C. I(+∞)=0=C ⇒I(s)=(1/2)ln∣((s^2 −25)/(s^2 −4))∣](https://www.tinkutara.com/question/Q150274.png)
$${I}\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{st}} \left(\mathrm{cosh}\left(\mathrm{2}{t}\right)−\mathrm{cosh}\left(\mathrm{5}{t}\right)\right)}{{t}}{dt} \\ $$$${I}\:'\left({s}\right)=−\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} \left(\mathrm{cosh}\left(\mathrm{2}{t}\right)−\mathrm{cosh}\left(\mathrm{5}{t}\right)\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} \left({e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} −{e}^{\mathrm{5}{t}} −{e}^{−\mathrm{5}{t}} \right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{e}^{−\left({s}−\mathrm{2}\right){t}} }{\mathrm{2}−{s}}−\frac{{e}^{−\left({s}+\mathrm{2}\right){t}} }{{s}+\mathrm{2}}−\frac{{e}^{−\left({s}−\mathrm{5}\right){t}} }{\mathrm{5}−{s}}+\frac{{e}^{−\left({s}+\mathrm{5}\right){t}} }{{s}+\mathrm{5}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}−{s}}−\frac{\mathrm{1}}{{s}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{5}−{s}}+\frac{\mathrm{1}}{{s}+\mathrm{5}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{I}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\mid{s}+\mathrm{5}\mid+\mathrm{ln}\mid{s}−\mathrm{5}\mid−\mathrm{ln}\mid{s}+\mathrm{2}\mid−\mathrm{ln}\mid{s}−\mathrm{2}\mid\right)+{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{{s}^{\mathrm{2}} −\mathrm{25}}{{s}^{\mathrm{2}} −\mathrm{4}}\mid+{C}.\:{I}\left(+\infty\right)=\mathrm{0}={C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{I}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{{s}^{\mathrm{2}} −\mathrm{25}}{{s}^{\mathrm{2}} −\mathrm{4}}\mid \\ $$
Commented by mathdanisur last updated on 10/Aug/21
$$\mathrm{Cool}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{Thank}\:{Y}\mathrm{ou} \\ $$
Commented by Ar Brandon last updated on 10/Aug/21
$$\mathcal{L}^{−\mathrm{1}} \left\{\frac{{s}+\mathrm{1}}{{s}^{\mathrm{2}} −\mathrm{16}{s}+\mathrm{100}}\right\}=\mathcal{L}^{−\mathrm{1}} \left\{\frac{{s}+\mathrm{1}}{\left({s}−\mathrm{8}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }\right\} \\ $$$$=\mathcal{L}^{−\mathrm{1}} \left\{\frac{{s}−\mathrm{8}}{\left({s}−\mathrm{8}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{9}}{\left({s}−\mathrm{8}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }\right\} \\ $$$$=\mathrm{cos}\left(\mathrm{6}\left({s}−\mathrm{8}\right)\right)+\frac{\mathrm{9}}{\mathrm{6}}\mathrm{sin}\left(\mathrm{6}\left({s}−\mathrm{8}\right)\right) \\ $$
Commented by Ar Brandon last updated on 10/Aug/21
$$\mathrm{You}\:\mathrm{deleted}\:\mathrm{your}\:\mathrm{previous}\:\mathrm{post}. \\ $$
Commented by mathdanisur last updated on 10/Aug/21
$$\mathrm{Cool}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{Thank}\:\mathrm{You} \\ $$