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Question Number 128893 by BHOOPENDRA last updated on 11/Jan/21

Answered by Dwaipayan Shikari last updated on 11/Jan/21

(1/2)∫_0 ^∞ e^(−3t) t^2 e^(2t) −e^(−3t−2t) t^2 dt          sinh2t=((e^(2t) −e^(−2t) )/2)  =(1/2)∫_0 ^∞ e^(−t) t^2 dt −(1/2)∫_0 ^∞ e^(−5t) t^2 dt  =(1/2)Γ(3)−(1/(5^3 .2))Γ(3)=((124)/(125))

$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{3}{t}} {t}^{\mathrm{2}} {e}^{\mathrm{2}{t}} −{e}^{−\mathrm{3}{t}−\mathrm{2}{t}} {t}^{\mathrm{2}} {dt}\:\:\:\:\:\:\:\:\:\:{sinh}\mathrm{2}{t}=\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {t}^{\mathrm{2}} {dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{5}{t}} {t}^{\mathrm{2}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} .\mathrm{2}}\Gamma\left(\mathrm{3}\right)=\frac{\mathrm{124}}{\mathrm{125}} \\ $$

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