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Question Number 219520 by OmoloyeMichael last updated on 27/Apr/25
find the laplace transform of  f(t)=∫_0 ^t ((sint)/t)dt
$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\int_{\mathrm{0}} ^{\boldsymbol{{t}}} \frac{\boldsymbol{{sint}}}{\boldsymbol{{t}}}\boldsymbol{{dt}} \\ $$
Answered by SdC355 last updated on 27/Apr/25
∫_0 ^( T)   ((sin(t))/t) dt=Si(T)  ∫_0 ^( ∞)   ((sin(t))/t)e^(−st)  dt=∫_s ^( ∞) ∫_0 ^( ∞)  sin(t)e^(−xt) dtdx  ∫_( s) ^( ∞)   (dx/(x^2 +1))=(π/2)−tan^(−1) (s)
$$\int_{\mathrm{0}} ^{\:{T}} \:\:\frac{\mathrm{sin}\left({t}\right)}{{t}}\:\mathrm{d}{t}=\mathrm{Si}\left({T}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}\left({t}\right)}{{t}}{e}^{−{st}} \:\mathrm{d}{t}=\int_{{s}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}\left({t}\right){e}^{−{xt}} \mathrm{d}{t}\mathrm{d}{x} \\ $$$$\int_{\:{s}} ^{\:\infty} \:\:\frac{\mathrm{d}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left({s}\right) \\ $$
Commented by OmoloyeMichael last updated on 28/Apr/25
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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