Question Number 16493 by Sai dadon. last updated on 23/Jun/17
$${Find}\:{L}\left\{{tcosh}\mathrm{2}{t}\right\} \\ $$
Commented by prakash jain last updated on 23/Jun/17
![∫_0 ^∞ tcosh 2te^(−st) dt =∫_0 ^( ∞) t((e^(2t) +e^(−2t) )/2)e^(−st) dt =(1/2)[∫_0 ^( ∞) te^((2−s)t) dt+∫_0 ^∞ te^(−(s+2)t) dt] =(1/2)[((te^((2−s)t) )/(2−s))−(e^((2−s)t) /((2−s)^2 ))−((te^(−(2+s)t) )/(2+s))−(e^(−(s+2)t) /((s+2)^2 ))]_0 ^∞ =(1/2)[(1/((2−s)^2 ))+(1/((2+s)^2 ))] =(((2+s)^2 +(2−s)^2 )/(2(s^2 −4)^2 ))=((s^2 +4)/((s^2 −4)^2 ))](https://www.tinkutara.com/question/Q16538.png)
$$\int_{\mathrm{0}} ^{\infty} {t}\mathrm{cosh}\:\mathrm{2}{te}^{−{st}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} {t}\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} }{\mathrm{2}}{e}^{−{st}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\:\infty} {te}^{\left(\mathrm{2}−{s}\right){t}} {dt}+\int_{\mathrm{0}} ^{\infty} {te}^{−\left({s}+\mathrm{2}\right){t}} {dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{te}^{\left(\mathrm{2}−{s}\right){t}} }{\mathrm{2}−{s}}−\frac{{e}^{\left(\mathrm{2}−{s}\right){t}} }{\left(\mathrm{2}−{s}\right)^{\mathrm{2}} }−\frac{{te}^{−\left(\mathrm{2}+{s}\right){t}} }{\mathrm{2}+{s}}−\frac{{e}^{−\left({s}+\mathrm{2}\right){t}} }{\left({s}+\mathrm{2}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\left(\mathrm{2}−{s}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}+{s}\right)^{\mathrm{2}} }\right] \\ $$$$=\frac{\left(\mathrm{2}+{s}\right)^{\mathrm{2}} +\left(\mathrm{2}−{s}\right)^{\mathrm{2}} }{\mathrm{2}\left({s}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }=\frac{{s}^{\mathrm{2}} +\mathrm{4}}{\left({s}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} } \\ $$