Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 167647 by mathlove last updated on 22/Mar/22

Answered by RoswelCod2003 last updated on 22/Mar/22

Using “Feynman′s Rule of Integration”     I(k) = ∫_0 ^( ∞) e^(−x^2 ) cos (kx) dx ∴ where k = 2  (d/dk) I(k) = (d/dk) ∫_0 ^( ∞) e^(−x^2 ) cos (kx) dx  (dI/dk) = ∫_0 ^( ∞) (∂/∂k) (e^(−x^2 ) cos (kx)) dx   (dI/dk) = ∫_0 ^( ∞) xe^(−x^2 ) sin (kx) dx ∴  determinant (((u = sin (kx) ; dv = −xe^(−x^2 ) )),((du = − k cos (kx) ; v = (1/2) e^(−x^2 ) )))  lim_(x→∞)  ((1/2) e^(−x^2 )  sin(kx)  ) − lim_(x→0)  ((1/2) e^(−x^2 ) sin(kx)) −(k/2)I(k)  − (k/2) I → (dI/dk) = −(k/2) I  (dI/I) = −(k/2) dk → ∫ (dI/I) = ∫ −(k/2) dk  ln ∣I∣ = −(k^2 /4) + C_1  → e^(ln∣I∣)  = e^(−(k^2 /4) + C_1 )   I(k) = C_2 e^(−(k^2 /4))   ∫_0 ^( ∞) e^(−x^2 ) cos (kx) dx = C_2 e^(−(k^2 /4))  ∴ let k = 0  ∫_0 ^( ∞) e^(−x^2 ) dx → ((√π)/2) = C_2  → ((√π)/2) e^(−(k^2 /4) )  ∴ k = 2  ∫_0 ^( ∞) e^(−x^2 ) cos (kx) dx = ((√π)/(2 )) e^(− 1)  or  ((√π)/(2e))     Final Answer:   ∫_0 ^( ∞) e^(−x^2 ) cos (2x) dx = ((√π)/(2e))     Roswel Bernardo :)

$$\mathrm{Using}\:``\mathrm{Feynman}'\mathrm{s}\:\mathrm{Rule}\:\mathrm{of}\:\mathrm{Integration}'' \\ $$$$\: \\ $$$${I}\left({k}\right)\:=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({kx}\right)\:{dx}\:\therefore\:\mathrm{where}\:{k}\:=\:\mathrm{2} \\ $$$$\frac{{d}}{{dk}}\:{I}\left({k}\right)\:=\:\frac{{d}}{{dk}}\:\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({kx}\right)\:{dx} \\ $$$$\frac{{dI}}{{dk}}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\partial}{\partial{k}}\:\left({e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({kx}\right)\right)\:{dx}\: \\ $$$$\frac{{dI}}{{dk}}\:=\:\int_{\mathrm{0}} ^{\:\infty} {xe}^{−{x}^{\mathrm{2}} } \mathrm{sin}\:\left({kx}\right)\:{dx}\:\therefore\:\begin{vmatrix}{{u}\:=\:\mathrm{sin}\:\left({kx}\right)\:;\:{dv}\:=\:−{xe}^{−{x}^{\mathrm{2}} } }\\{{du}\:=\:−\:{k}\:\mathrm{cos}\:\left({kx}\right)\:;\:{v}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{x}^{\mathrm{2}} } }\end{vmatrix} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{x}^{\mathrm{2}} } \:\mathrm{sin}\left({kx}\right)\:\:\right)\:−\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{x}^{\mathrm{2}} } \mathrm{sin}\left({kx}\right)\right)\:−\frac{{k}}{\mathrm{2}}{I}\left({k}\right) \\ $$$$−\:\frac{{k}}{\mathrm{2}}\:{I}\:\rightarrow\:\frac{{dI}}{{dk}}\:=\:−\frac{{k}}{\mathrm{2}}\:{I} \\ $$$$\frac{{dI}}{{I}}\:=\:−\frac{{k}}{\mathrm{2}}\:{dk}\:\rightarrow\:\int\:\frac{{dI}}{{I}}\:=\:\int\:−\frac{{k}}{\mathrm{2}}\:{dk} \\ $$$$\mathrm{ln}\:\mid{I}\mid\:=\:−\frac{{k}^{\mathrm{2}} }{\mathrm{4}}\:+\:{C}_{\mathrm{1}} \:\rightarrow\:{e}^{\mathrm{ln}\mid{I}\mid} \:=\:{e}^{−\frac{{k}^{\mathrm{2}} }{\mathrm{4}}\:+\:{C}_{\mathrm{1}} } \\ $$$${I}\left({k}\right)\:=\:{C}_{\mathrm{2}} {e}^{−\frac{{k}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({kx}\right)\:{dx}\:=\:{C}_{\mathrm{2}} {e}^{−\frac{{k}^{\mathrm{2}} }{\mathrm{4}}} \:\therefore\:{let}\:{k}\:=\:\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } {dx}\:\rightarrow\:\frac{\sqrt{\pi}}{\mathrm{2}}\:=\:{C}_{\mathrm{2}} \:\rightarrow\:\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{−\frac{{k}^{\mathrm{2}} }{\mathrm{4}}\:} \:\therefore\:{k}\:=\:\mathrm{2} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({kx}\right)\:{dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}\:}\:{e}^{−\:\mathrm{1}} \:{or}\:\:\frac{\sqrt{\pi}}{\mathrm{2}{e}} \\ $$$$\: \\ $$$${Final}\:{Answer}:\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left(\mathrm{2}{x}\right)\:{dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}{e}} \\ $$$$\: \\ $$$$\left.{Roswel}\:{Bernardo}\::\right) \\ $$$$\: \\ $$

Commented by mkam last updated on 22/Mar/22

sir can you given me the Feymans Rule formulla

$${sir}\:{can}\:{you}\:{given}\:{me}\:{the}\:{Feymans}\:{Rule}\:{formulla} \\ $$

Commented by mkam last updated on 22/Mar/22

sir can you given me the Feymans Rule formulla

$${sir}\:{can}\:{you}\:{given}\:{me}\:{the}\:{Feymans}\:{Rule}\:{formulla} \\ $$

Commented by RoswelCod2003 last updated on 22/Mar/22

   Ahmmm... The Feynman′s Rule or Feynman′s Trick  is performed by differentiating undwr the integral sign.      Example:     ∫_(−1) ^( 1) (2kx + k^2 ) dx  set this integral as function of any variable, let′s “k”  F(k) = ∫_(−1) ^( 1) (2kx + k^2 ) dx  get the derivative both sides with respect to k    (d/dk) F(k) = (d/dk)∫_(−1) ^( 1) (2kx + k^2 ) dx ∴ the derivative of F(k) is simply F ′(k) or (dF/dk)  (dF/dk) = (d/dk) ∫_(−1) ^( 1) (2kx + k^2 ) dx ∴ before you get the derivative of (2kx + k^2 )  put first the derivative notion inside the integral sign but transform it  into partial derivative notation:     (dF/dk) = ∫_(−1) ^( 1) (∂/∂k)(2kx + k^2 ) dx ∴ then the derivative of the function under the  integral sign.     we will have: (dF/dk) = ∫_(−1) ^( 1) (2x + 2k) dx ∴ now integrate with  respect to X:     (dF/dk) =  ((2x^(1+1) )/(1+1)) → ((2x^2 )/2) = x^2  + 2k       2k will remain to 2k because we are integrating  with respect to x not k     we will have: [x^2  + 2k]_(−1) ^1   apply now the fundamental theorem of calculus     ∫_a ^( b) f(x) dx  = f(b) − f(a)  [f(b) =  (x^2  + 2k)] − [f(a) = (x^2  + 2k)]  [f(1) = (1 + 2k)] −[f(−1) = (1 + 2k)]  1 + 2k − 1 + 2k = 4k     Final Answer:   ∫_(−1) ^( 1) (2kx + k^2 ) dx = 4k     hope you learned something, you can actually   learn this from youtube:)     Roswel Bernardo at your service <3

$$\: \\ $$$${Ahmmm}...\:{The}\:{Feynman}'{s}\:{Rule}\:{or}\:{Feynman}'{s}\:{Trick} \\ $$$${is}\:{performed}\:{by}\:{differentiating}\:{undwr}\:{the}\:{integral}\:{sign}.\: \\ $$$$\: \\ $$$$\mathrm{E}{xample}: \\ $$$$\: \\ $$$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left(\mathrm{2}{kx}\:+\:{k}^{\mathrm{2}} \right)\:{dx} \\ $$$${set}\:{this}\:{integral}\:{as}\:{function}\:{of}\:{any}\:{variable},\:{let}'{s}\:``{k}'' \\ $$$${F}\left({k}\right)\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left(\mathrm{2}{kx}\:+\:{k}^{\mathrm{2}} \right)\:{dx} \\ $$$${get}\:{the}\:{derivative}\:{both}\:{sides}\:{with}\:{respect}\:{to}\:{k} \\ $$$$ \\ $$$$\frac{{d}}{{dk}}\:{F}\left({k}\right)\:=\:\frac{{d}}{{dk}}\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left(\mathrm{2}{kx}\:+\:{k}^{\mathrm{2}} \right)\:{dx}\:\therefore\:{the}\:{derivative}\:{of}\:{F}\left({k}\right)\:{is}\:{simply}\:{F}\:'\left({k}\right)\:{or}\:\frac{{dF}}{{dk}} \\ $$$$\frac{{dF}}{{dk}}\:=\:\frac{{d}}{{dk}}\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left(\mathrm{2}{kx}\:+\:{k}^{\mathrm{2}} \right)\:{dx}\:\therefore\:{before}\:{you}\:{get}\:{the}\:{derivative}\:{of}\:\left(\mathrm{2}{kx}\:+\:{k}^{\mathrm{2}} \right) \\ $$$${put}\:{first}\:{the}\:{derivative}\:{notion}\:{inside}\:{the}\:{integral}\:{sign}\:{but}\:{transform}\:{it} \\ $$$${into}\:\boldsymbol{\mathrm{partial}}\:\boldsymbol{\mathrm{derivative}}\:\boldsymbol{\mathrm{notation}}: \\ $$$$\: \\ $$$$\frac{{dF}}{{dk}}\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \frac{\partial}{\partial{k}}\left(\mathrm{2}{kx}\:+\:{k}^{\mathrm{2}} \right)\:{dx}\:\therefore\:{then}\:{the}\:{derivative}\:{of}\:{the}\:{function}\:{under}\:{the} \\ $$$${integral}\:{sign}. \\ $$$$\: \\ $$$${we}\:{will}\:{have}:\:\frac{{dF}}{{dk}}\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left(\mathrm{2}{x}\:+\:\mathrm{2}{k}\right)\:{dx}\:\therefore\:{now}\:{integrate}\:{with} \\ $$$${respect}\:{to}\:{X}: \\ $$$$\: \\ $$$$\frac{{dF}}{{dk}}\:=\:\:\frac{\mathrm{2}{x}^{\mathrm{1}+\mathrm{1}} }{\mathrm{1}+\mathrm{1}}\:\rightarrow\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{k} \\ $$$$\: \\ $$$$ \\ $$$$\mathrm{2}{k}\:{will}\:{remain}\:{to}\:\mathrm{2}{k}\:{because}\:{we}\:{are}\:{integrating} \\ $$$${with}\:{respect}\:{to}\:{x}\:{not}\:{k} \\ $$$$\: \\ $$$${we}\:{will}\:{have}:\:\left[{x}^{\mathrm{2}} \:+\:\mathrm{2}{k}\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$${apply}\:{now}\:{the}\:{fundamental}\:{theorem}\:{of}\:{calculus} \\ $$$$\: \\ $$$$\int_{{a}} ^{\:{b}} {f}\left({x}\right)\:{dx}\:\:=\:{f}\left({b}\right)\:−\:{f}\left({a}\right) \\ $$$$\left[{f}\left({b}\right)\:=\:\:\left({x}^{\mathrm{2}} \:+\:\mathrm{2}{k}\right)\right]\:−\:\left[{f}\left({a}\right)\:=\:\left({x}^{\mathrm{2}} \:+\:\mathrm{2}{k}\right)\right] \\ $$$$\left[{f}\left(\mathrm{1}\right)\:=\:\left(\mathrm{1}\:+\:\mathrm{2}{k}\right)\right]\:−\left[{f}\left(−\mathrm{1}\right)\:=\:\left(\mathrm{1}\:+\:\mathrm{2}{k}\right)\right] \\ $$$$\mathrm{1}\:+\:\mathrm{2}{k}\:−\:\mathrm{1}\:+\:\mathrm{2}{k}\:=\:\mathrm{4}{k} \\ $$$$\: \\ $$$$\mathrm{Final}\:\mathrm{Answer}:\: \\ $$$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left(\mathrm{2}{kx}\:+\:{k}^{\mathrm{2}} \right)\:{dx}\:=\:\mathrm{4}{k} \\ $$$$\: \\ $$$${hope}\:{you}\:{learned}\:{something},\:{you}\:{can}\:{actually}\: \\ $$$$\left.{learn}\:{this}\:{from}\:{youtube}:\right) \\ $$$$\: \\ $$$$\mathrm{Roswel}\:\mathrm{Bernardo}\:\mathrm{at}\:\mathrm{your}\:\mathrm{service}\:<\mathrm{3} \\ $$

Answered by Mathspace last updated on 22/Mar/22

∫_0 ^∞  e^(−x^2 ) cos(2x)dx  =(1/2)∫_(−∞) ^(+∞)  e^(−x^2 ) cos(2x)dx  =(1/2)Re(∫_(−∞) ^(+∞)  e^(−x^2 +2ix) dx) and  ∫_(−∞) ^(+∞)  e^(−x^2 +2ix ) dx=∫_(−∞) ^(+∞)  e^(−(x^2 −2ix +i^2 −i^2 )) dx  =e^(−1) ∫_(−∞) ^(+∞)  e^(−(x−i)^2 )   dx (x−i=t)  =e^(−1) ∫_(−∞) ^(+∞)  e^(−t^2 ) dt  =e^(−1) .(√π) ⇒  ∫_0 ^∞   e^(−x^2 ) cos(2x)dx=((e^(−1) (√π))/2)  =((√π)/(2e))

$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {cos}\left(\mathrm{2}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {cos}\left(\mathrm{2}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{Re}\left(\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} +\mathrm{2}{ix}} {dx}\right)\:{and} \\ $$$$\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} +\mathrm{2}{ix}\:} {dx}=\int_{−\infty} ^{+\infty} \:{e}^{−\left({x}^{\mathrm{2}} −\mathrm{2}{ix}\:+{i}^{\mathrm{2}} −{i}^{\mathrm{2}} \right)} {dx} \\ $$$$={e}^{−\mathrm{1}} \int_{−\infty} ^{+\infty} \:{e}^{−\left({x}−{i}\right)^{\mathrm{2}} } \:\:{dx}\:\left({x}−{i}={t}\right) \\ $$$$={e}^{−\mathrm{1}} \int_{−\infty} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$={e}^{−\mathrm{1}} .\sqrt{\pi}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}^{\mathrm{2}} } {cos}\left(\mathrm{2}{x}\right){dx}=\frac{{e}^{−\mathrm{1}} \sqrt{\pi}}{\mathrm{2}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}{e}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com