Question Number 82244 by M±th+et£s last updated on 19/Feb/20
![find the function of f when this function continue at interval [−∞,0] ∫_(−x^2 ) ^0 f(t) dt=(d/dx)[x(1−sin(πx)]](https://www.tinkutara.com/question/Q82244.png)
$${find}\:{the}\:{function}\:{of}\:{f}\:{when}\:{this}\:\: \\ $$$${function}\:{continue}\:{at}\:{interval}\:\left[−\infty,\mathrm{0}\right] \\ $$$$\int_{−{x}^{\mathrm{2}} } ^{\mathrm{0}} {f}\left({t}\right)\:{dt}=\frac{{d}}{{dx}}\left[{x}\left(\mathrm{1}−{sin}\left(\pi{x}\right)\right]\right. \\ $$
Commented by mr W last updated on 19/Feb/20
![∫_(−x^2 ) ^0 f(t) dt=(d/dx)[x(1−sin(πx)] ∫_(−x^2 ) ^0 f(t) dt=1−sin(πx)−πx cos (πx) 2xf(−x^2 )=−2πcos (πx)+π^2 x sin (πx) f(−x^2 )=−(π/x)cos (πx)+(π^2 /2) sin (πx) t=−x^2 ⇒x=±(√(−t)) f(t)=∓(π/( (√(−t))))cos (π(√(−t)))±(π^2 /2) sin (π(√(−t))) ⇒f(x)=∓(π/( (√(−x))))cos (π(√(−x)))±(π^2 /2) sin (π(√(−x)))](https://www.tinkutara.com/question/Q82252.png)
$$\int_{−{x}^{\mathrm{2}} } ^{\mathrm{0}} {f}\left({t}\right)\:{dt}=\frac{{d}}{{dx}}\left[{x}\left(\mathrm{1}−{sin}\left(\pi{x}\right)\right]\right. \\ $$$$\int_{−{x}^{\mathrm{2}} } ^{\mathrm{0}} {f}\left({t}\right)\:{dt}=\mathrm{1}−{sin}\left(\pi{x}\right)−\pi{x}\:\mathrm{cos}\:\left(\pi{x}\right) \\ $$$$\mathrm{2}{xf}\left(−{x}^{\mathrm{2}} \right)=−\mathrm{2}\pi\mathrm{cos}\:\left(\pi{x}\right)+\pi^{\mathrm{2}} {x}\:\mathrm{sin}\:\left(\pi{x}\right) \\ $$$${f}\left(−{x}^{\mathrm{2}} \right)=−\frac{\pi}{{x}}\mathrm{cos}\:\left(\pi{x}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{sin}\:\left(\pi{x}\right) \\ $$$${t}=−{x}^{\mathrm{2}} \:\Rightarrow{x}=\pm\sqrt{−{t}} \\ $$$${f}\left({t}\right)=\mp\frac{\pi}{\:\sqrt{−{t}}}\mathrm{cos}\:\left(\pi\sqrt{−{t}}\right)\pm\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{sin}\:\left(\pi\sqrt{−{t}}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\mp\frac{\pi}{\:\sqrt{−{x}}}\mathrm{cos}\:\left(\pi\sqrt{−{x}}\right)\pm\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{sin}\:\left(\pi\sqrt{−{x}}\right) \\ $$
Commented by M±th+et£s last updated on 19/Feb/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Answered by mind is power last updated on 19/Feb/20
![⇒2xf(−x^2 )=(d^2 /dx^2 )[x−xsin(πx)] ⇒f(−x^2 )=(1/(2x))(d/dx)[−sin(πx)−πxcos(πx)] ⇒f(−x^2 )=(1/(2x)).[−2πcos(πx)+π^2 xsin(πx)] f(−x^2 )=−((πcos(πx))/x)+(π^2 /2)sin(πx) −x^2 =y⇒x=+_− (√(−y)) we got 2 solution f(y)=−((πcos(π(√(−y))))/( (√(−y))))+((π^2 sin(π(√(−y))))/2) y≠0 f(0)=0 and −f](https://www.tinkutara.com/question/Q82260.png)
$$\Rightarrow\mathrm{2}{xf}\left(−{x}^{\mathrm{2}} \right)=\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left[{x}−{xsin}\left(\pi{x}\right)\right] \\ $$$$\Rightarrow{f}\left(−{x}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}{x}}\frac{{d}}{{dx}}\left[−{sin}\left(\pi{x}\right)−\pi{xcos}\left(\pi{x}\right)\right] \\ $$$$\Rightarrow{f}\left(−{x}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}{x}}.\left[−\mathrm{2}\pi{cos}\left(\pi{x}\right)+\pi^{\mathrm{2}} {xsin}\left(\pi{x}\right)\right] \\ $$$${f}\left(−{x}^{\mathrm{2}} \right)=−\frac{\pi{cos}\left(\pi{x}\right)}{{x}}+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{sin}\left(\pi{x}\right) \\ $$$$−{x}^{\mathrm{2}} ={y}\Rightarrow{x}=\underset{−} {+}\sqrt{−{y}} \\ $$$${we}\:{got}\:\mathrm{2}\:{solution}\: \\ $$$${f}\left({y}\right)=−\frac{\pi{cos}\left(\pi\sqrt{−{y}}\right)}{\:\sqrt{−{y}}}+\frac{\pi^{\mathrm{2}} {sin}\left(\pi\sqrt{−{y}}\right)}{\mathrm{2}}\:\:{y}\neq\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$ \\ $$$${and}\:−{f} \\ $$
Commented by M±th+et£s last updated on 19/Feb/20
$${god}\:{bless}\:{you}\:{sir} \\ $$