Question Number 205599 by mnjuly1970 last updated on 25/Mar/24
$$ \\ $$$$\:{laplace}\:{transform}... \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\mathscr{L}\:\left\{\:\:{sin}\left(\sqrt{{t}}\:\right)\right\}\:=? \\ $$$$−−−− \\ $$
Commented by SANOGO last updated on 25/Mar/24
$${merci}\:{beaucoup} \\ $$
Answered by Berbere last updated on 25/Mar/24
![L{sin((√t))}(x)=∫_0 ^∞ sin((√t))e^(−xt) dt;x>0 (√t)=y =∫_0 ^∞ sin(y)e^(−xy^2 ) 2ydy=^(IBP) [−(e^(−xy^2 ) /x)sin(y)]_0 ^∞ +(1/x)∫_0 ^∞ cos(y)e^(−xy^2 ) dy =(1/x)Re∫_0 ^∞ e^(−xy^2 +iy) dy=(1/x)Re∫_0 ^∞ e^(−(y(√x)+(i/(2(√x))))^2 −(1/(4x))) dy y(√x).=u,(e^(−(1/(4x))) /(x(√x)))Re∫_0 ^∞ e^(−(u+(i/(2(√x))))^2 ) du X=u+(i/(2(√x)));(e^(−(1/(4x))) /(x(√x)))Re∫_(0+(i/(2(√x)))) ^(∞+(i/(2(√x)))) e^(−X^2 ) dX; let rectangle ofDq (0,0) (0,(i/(2(√x))));(∞,(i/(2(√x))));(∞,0) f(z)=e^(−z^2 ) ∫_D f(z)dz=0 Holomorphic function cauchy Theorem ∫_0 ^(i/(2(√x))) f(z)dz+∫_(i/(2(√x))) ^(∞+(i/(2(√x)))) f(z)dz+∫_(∞+(i/(2(√x)))) ^∞ f(z)dz_(=0) +∫_∞ ^0 f(z)dz=0 z=x+iy ∣f(z)∣≤e^(−y^2 ) .e^(−x^2 ) →0 x→∞ ∫_0 ^∞ e^(−z^2 ) =(1/2)Γ((1/2))=((√π)/2);∫_0 ^(i/(2(√x))) e^(−z^2 ) dz;z=it =i∫_0 ^(1/(2(√x))) e^t^2 dt imaginair Pur Re∫_(0+(i/(2(√x)))) ^(∞+(i/(2(√x)))) e^(−X^2 ) =((√π)/2) L(sin((√t))(x)=(e^(−(1/(4x))) /(x(√x))).((√π)/2)](Q205601.png)
$$\mathscr{L}\left\{{sin}\left(\sqrt{{t}}\right)\right\}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {sin}\left(\sqrt{{t}}\right){e}^{−{xt}} {dt};{x}>\mathrm{0} \\ $$$$\sqrt{{t}}={y} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {sin}\left({y}\right){e}^{−{xy}^{\mathrm{2}} } \mathrm{2}{ydy}\overset{{IBP}} {=}\left[−\frac{{e}^{−{xy}^{\mathrm{2}} } }{{x}}{sin}\left({y}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} {cos}\left({y}\right){e}^{−{xy}^{\mathrm{2}} } {dy} \\ $$$$=\frac{\mathrm{1}}{{x}}{Re}\int_{\mathrm{0}} ^{\infty} {e}^{−{xy}^{\mathrm{2}} +{iy}} {dy}=\frac{\mathrm{1}}{{x}}{Re}\int_{\mathrm{0}} ^{\infty} {e}^{−\left({y}\sqrt{{x}}+\frac{{i}}{\mathrm{2}\sqrt{{x}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}{x}}} {dy} \\ $$$${y}\sqrt{{x}}.={u},\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{4}{x}}} }{{x}\sqrt{{x}}}{Re}\int_{\mathrm{0}} ^{\infty} {e}^{−\left({u}+\frac{{i}}{\mathrm{2}\sqrt{{x}}}\right)^{\mathrm{2}} } {du} \\ $$$${X}={u}+\frac{{i}}{\mathrm{2}\sqrt{{x}}};\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{4}{x}}} }{{x}\sqrt{{x}}}{Re}\int_{\mathrm{0}+\frac{{i}}{\mathrm{2}\sqrt{{x}}}} ^{\infty+\frac{{i}}{\mathrm{2}\sqrt{{x}}}} {e}^{−{X}^{\mathrm{2}} } {dX}; \\ $$$${let}\:{rectangle}\:{ofDq}\:\left(\mathrm{0},\mathrm{0}\right)\:\left(\mathrm{0},\frac{{i}}{\mathrm{2}\sqrt{{x}}}\right);\left(\infty,\frac{{i}}{\mathrm{2}\sqrt{{x}}}\right);\left(\infty,\mathrm{0}\right) \\ $$$${f}\left({z}\right)={e}^{−{z}^{\mathrm{2}} } \\ $$$$\int_{{D}} {f}\left({z}\right){dz}=\mathrm{0}\:\:{Holomorphic}\:{function}\:{cauchy}\:{Theorem} \\ $$$$\int_{\mathrm{0}} ^{\frac{{i}}{\mathrm{2}\sqrt{{x}}}} {f}\left({z}\right){dz}+\int_{\frac{{i}}{\mathrm{2}\sqrt{{x}}}} ^{\infty+\frac{{i}}{\mathrm{2}\sqrt{{x}}}} {f}\left({z}\right){dz}+\int_{\infty+\frac{{i}}{\mathrm{2}\sqrt{{x}}}} ^{\infty} {f}\left({z}\right){dz}_{=\mathrm{0}} +\int_{\infty} ^{\mathrm{0}} {f}\left({z}\right){dz}=\mathrm{0} \\ $$$${z}={x}+{iy}\:\:\mid{f}\left({z}\right)\mid\leqslant{e}^{−{y}^{\mathrm{2}} } .{e}^{−{x}^{\mathrm{2}} } \rightarrow\mathrm{0}\:{x}\rightarrow\infty \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{z}^{\mathrm{2}} } =\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}};\int_{\mathrm{0}} ^{\frac{{i}}{\mathrm{2}\sqrt{{x}}}} {e}^{−{z}^{\mathrm{2}} } {dz};{z}={it} \\ $$$$={i}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}} {e}^{{t}^{\mathrm{2}} } {dt}\:{imaginair}\:{Pur} \\ $$$${Re}\int_{\mathrm{0}+\frac{{i}}{\mathrm{2}\sqrt{{x}}}} ^{\infty+\frac{{i}}{\mathrm{2}\sqrt{{x}}}} {e}^{−{X}^{\mathrm{2}} } =\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathcal{L}}\left({sin}\left(\sqrt{{t}}\right)\left({x}\right)=\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{4}{x}}} }{{x}\sqrt{{x}}}.\frac{\sqrt{\pi}}{\mathrm{2}}\right. \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 25/Mar/24
$${thanks}\:{alot}\:\:{Master}\: \\ $$
Commented by Berbere last updated on 27/Mar/24
$${withe}\:{pleasur} \\ $$