Without using Laplace transforms, how do I show that for every positive number $x$ the following equation is valid? $$\int_{0}^{\infty}e^{-xt}\sin(t)dt=\frac{1}{x^2+1}. $$
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Yet another one (particularly useful when evaluating particular integral of ODE): Given integral = imaginary part of $\int_0^\infty e^{-xt}.e^{it}dt$ = imaginary part of $\int_0^\infty e^{-(x-i)t}dt$ = imaginary part of $\frac{1}{x-i}$ = imaginary part of $\frac{x+i}{(x-i)(x+i)}$ = $\frac{1}{1+x^2}$ |
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$\rm\bf Hint$: $$\large e^{it}=\cos(t)+i\sin(t)\implies \sin(t)=\frac{e^{it}-e^{-it}}{2i}.$$ |
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Why the "complex analysis" tag? It's just a double integration: $$ \int e^{-xt}\sin(t)dt=-\frac{1}{x}e^{-xt}\sin(t)+\frac{1}{x}\int e^{-xt}\cos(t)dt= $$ $$ =-\frac{1}{x}e^{-xt}\sin(t)-\frac{1}{x^2}e^{-xt}\cos(t)-\frac{1}{x^2}\int e^{-xt}\sin(t)dt $$ It follows that $$ \left(1+\frac{1}{x^2}\right)\int e^{-xt}\sin(t)dt= -\frac{1}{x}e^{-xt}\sin(t)-\frac{1}{x^2}e^{-xt}\cos(t)+ $$ It follows that $$ \left(1+\frac{1}{x^2}\right)\int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{x^2} $$ $$ \left(\frac{x^2+1}{x^2}\right)\int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{x^2} $$ $$ \int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{1+x^2} $$ It follows that the integral is equal to $\displaystyle \frac{1}{x^2+1}$, as you wrote. |
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If you integrate by parts twice, using $u=\exp(-xt), dv=\sin(t)dt$ then $dv= \cos(t)dt$ you get the same integral back and can solve for it. |
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