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if-a-gt-0-and-m-0-0-cos-mx-x-2-a-2-2-dx-

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Question Number 201341 by mathlove last updated on 05/Dec/23
if a>0 and m≥0 ∫_0 ^∞ ((cos(mx))/((x^2 +a^2 )^2 ))dx=?
$${if}\:{a}>\mathrm{0}\:\:{and}\:\:{m}\geqslant\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({mx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=? \\ $$
Answered by Calculusboy last updated on 05/Dec/23
Solution: since a>0 and m>_− 0 let a=1 and m=0 I=∫_0 ^∞ ((cos(0×x))/((x^2 +1)^2 ))dx ⇔ I=∫_0 ^∞ (1/((x^2 +1)^2 ))dx let x=tan𝛉 dx=sec^2 𝛉d𝛉 when x=∞ 𝛉=(𝛑/2) and when x=0 𝛉=0 I=∫_0 ^(𝛑/2) ((sec^2 𝛉d𝛉)/((tan^2 𝛉+1)^2 )) ⇔ I=∫_0 ^(𝛑/2) ((sec^2 𝛉d𝛉)/((sec^2 𝛉)^2 )) I=∫_0 ^(𝛑/2) ((sec^2 𝛉)/(sec^4 𝛉)) d𝛉 ⇔ I=∫_0 ^(𝛑/2) (1/(sec^2 𝛉))d𝛉 I=∫_0 ^(𝛑/2) cos^2 𝛉 d𝛉 by using wallis formula(the power is even) I_n =((n−1)/n)×(𝛑/2) (n=2) I=((2−1)/2)×(𝛑/2) I=(𝛑/4)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{since}}\:\boldsymbol{{a}}>\mathrm{0}\:\boldsymbol{{and}}\:\boldsymbol{{m}}\underset{−} {>}\mathrm{0}\:\:\:\boldsymbol{{let}}\:\boldsymbol{{a}}=\mathrm{1}\:\boldsymbol{{and}}\:\boldsymbol{{m}}=\mathrm{0} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{cos}}\left(\mathrm{0}×\boldsymbol{{x}}\right)}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\boldsymbol{{dx}}\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\boldsymbol{{dx}}\:\:\:\boldsymbol{{let}}\:\boldsymbol{{x}}=\boldsymbol{{tan}\theta}\:\:\:\boldsymbol{{dx}}=\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta} \\ $$$$\boldsymbol{{when}}\:\:\boldsymbol{{x}}=\infty\:\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{\theta}=\mathrm{0} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta}}{\left(\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{\theta}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta}}{\left(\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta}\right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta}}{\boldsymbol{{sec}}^{\mathrm{4}} \boldsymbol{\theta}}\:\boldsymbol{{d}\theta}\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta}}\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{\theta}\:\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{wallis}}\:\boldsymbol{{formula}}\left(\boldsymbol{{the}}\:\boldsymbol{{power}}\:\boldsymbol{{is}}\:\boldsymbol{{even}}\right) \\ $$$$\boldsymbol{{I}}_{\boldsymbol{{n}}} =\frac{\boldsymbol{{n}}−\mathrm{1}}{\boldsymbol{{n}}}×\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\left(\boldsymbol{{n}}=\mathrm{2}\right) \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{2}−\mathrm{1}}{\mathrm{2}}×\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$
Commented by mathlove last updated on 05/Dec/23
thanks
$${thanks} \\ $$
Answered by Mathspace last updated on 05/Dec/23
f(a)=∫_0 ^∞ ((cos(mx))/(x^2 +a^2 ))dx f^′ (a)=−2a∫_0 ^∞ ((cos(mx))/((x^2 +a^2 )^2 ))dx ⇒ ∫_0 ^∞ ((cos(mx))/((x^2 +a^2 )^2 ))dx=−(1/(2a))f^′ (a) but f(a)=Re((1/2)∫_R (e^(imx) /(x^2 +a^2 ))dx) f(z)=(e^(imz) /(z^2 +a^2 ))=(e^(imz) /((z−ia)(z+ia))) the poles of f are ia and −ia ∫_R f(z)dz=2iπΣ_(ai ∈p^+ ) Resa_i =2iπRes(f,ia) =2iπ.(e^(im(ia)) /(2ia))=(π/a)e^(−ma) ⇒f(a)=(π/(2a))e^(−ma) and f^′ (a)=−(π/(2a^2 ))e^(−ma) −((mπ)/a)e^(−ma) finally ∫_0 ^∞ ((cos(mx))/((x^2 +a^2 )))dx =−(1/(2a)){−(π/a^2 )−((mπ)/a)}e^(−ma) =((π/(2a^3 ))+((ma)/(2a^2 )))e^(−ma)
$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({mx}\right)}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx} \\ $$$${f}^{'} \left({a}\right)=−\mathrm{2}{a}\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({mx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({mx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{\mathrm{2}{a}}{f}^{'} \left({a}\right) \\ $$$${but}\:{f}\left({a}\right)={Re}\left(\frac{\mathrm{1}}{\mathrm{2}}\int_{{R}} \:\frac{{e}^{{imx}} }{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}\right) \\ $$$${f}\left({z}\right)=\frac{{e}^{{imz}} }{{z}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{e}^{{imz}} }{\left({z}−{ia}\right)\left({z}+{ia}\right)} \\ $$$${the}\:{poles}\:{of}\:{f}\:{are}\:{ia}\:{and}\:−{ia} \\ $$$$\int_{{R}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi\sum_{{ai}\:\in{p}^{+} } {Resa}_{{i}} \\ $$$$=\mathrm{2}{i}\pi{Res}\left({f},{ia}\right) \\ $$$$=\mathrm{2}{i}\pi.\frac{{e}^{{im}\left({ia}\right)} }{\mathrm{2}{ia}}=\frac{\pi}{{a}}{e}^{−{ma}} \:\Rightarrow{f}\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}{e}^{−{ma}} \\ $$$${and}\:{f}^{'} \left({a}\right)=−\frac{\pi}{\mathrm{2}{a}^{\mathrm{2}} }{e}^{−{ma}} −\frac{{m}\pi}{{a}}{e}^{−{ma}} \\ $$$${finally}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({mx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{a}}\left\{−\frac{\pi}{{a}^{\mathrm{2}} }−\frac{{m}\pi}{{a}}\right\}{e}^{−{ma}} \\ $$$$=\left(\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }+\frac{{ma}}{\mathrm{2}{a}^{\mathrm{2}} }\right){e}^{−{ma}} \\ $$$$ \\ $$
Commented by Mathspace last updated on 05/Dec/23
sorry ∫_0 ^∞ ((cos(mx))/((x^2 +a^2 )^2 ))dx =−(1/(2a)){−(π/(2a^2 ))e^(−ma) −((mπ)/(2a))}e^(−ma) =((π/(4a^3 ))+((mπ)/(4a^2 )))e^(−ma) =(π/(4a^3 ))(1+am)e^(−ma)
$${sorry}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({mx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{a}}\left\{−\frac{\pi}{\mathrm{2}{a}^{\mathrm{2}} }{e}^{−{ma}} −\frac{{m}\pi}{\mathrm{2}{a}}\right\}{e}^{−{ma}} \\ $$$$=\left(\frac{\pi}{\mathrm{4}{a}^{\mathrm{3}} }+\frac{{m}\pi}{\mathrm{4}{a}^{\mathrm{2}} }\right){e}^{−{ma}} \\ $$$$=\frac{\pi}{\mathrm{4}{a}^{\mathrm{3}} }\left(\mathrm{1}+{am}\right){e}^{−{ma}} \\ $$

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