Menu Close

Question-159552

Tinku Tara 0 Comments
Pin




Question Number 159552 by Ar Brandon last updated on 18/Nov/21
Commented by Ar Brandon last updated on 18/Nov/21
Prove the above results
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{above}\:\mathrm{results} \\ $$
Commented by mindispower last updated on 19/Nov/21
bonjour tes Etudiant?
$${bonjour}\:{tes}\:{Etudiant}? \\ $$
Commented by Ar Brandon last updated on 19/Nov/21
Oui monsieur. E^� tudiant en 1^(er) anne^� e IUT.
$$\mathrm{Oui}\:\mathrm{monsieur}.\:\acute {\mathrm{E}tudiant}\:\mathrm{en}\:\mathrm{1}^{\mathrm{er}} \:\mathrm{ann}\acute {\mathrm{e}e}\:\mathrm{IUT}. \\ $$
Commented by puissant last updated on 20/Nov/21
��������
Answered by mindispower last updated on 19/Nov/21
(1) let f(b)=∫_0 ^∞ ((cos(ax)ln(x^2 +bz^2 ))/(x^2 +z^2 )),b≥0 f′(b)=∫_0 ^∞ ((z^2 cos(ax))/((x^2 +z^2 )(x^2 +bz^2 )))dx =z^2 ∫_0 ^∞ ((cos(ax))/((x^2 +z^2 )(x^2 +bz^2 )))dx=(z^2 /2)∫_(−∞) ^∞ ((cos(ax))/((x^2 +z^2 )(x^2 +bz^2 )))dx A=(z^2 /2)∫_(−∞) ^∞ (e^(iax) /((x^2 +z^2 )(x^2 +bz^2 )))dx= =z^2 .iπ.(e^(−az) /(2iz)).(1/((1−b)z^2 ))+iπ.(e^(−az(√b)) /(z^2 (1−b).(2iz(√b)))) =(π/(2z))((e^(−az) /((1−b)))+(e^(−az(√b)) /((1−b)(√b)))) f(b)=(π/(2z))∫(e^(−az) /(1−b))+(e^(−az(√b)) /((1−b)(√b)))db =(π/(2z))(−ln(1−b)e^(−az) +∫(e^(−az(√b)) /((1−b)(√b)))db∣_((√b)=u_ ) ^ ∫(e^(−auz) /((1−u^2 ))).2du =∫(e^(−auz) /(1−u))+(e^(−auz) /(1+u))du =−∫(e^(−az(1−x)) /x)dx+∫(e^(−az(y−1)) /y)dy =e^(−az) .−∫(e^(azx) /(azx))dazx+e^(az) ∫(e^(−azy) /(azy))d(azy) =e^(−az) −E_1 (−azx)+e^(az) .E_1 (azy) =−e^(−az) E_1 (−az(1−(√b)))+e^(az) (az(1+(√b))) −e^(az) .E_i (−az(1+(√b))+e^(−az) E_i (az(1−(√b))) we get (π/(2z))(e^(−az) (E_i (az(1−(√b))−ln(1−b))−e^(az) E_i (−az(1+(√b))))+C f(0)=∫_0 ^∞ ((cos(ax)ln(x^2 ))/(x^2 +z^2 ))dx=2∫_0 ^∞ ((cos(ax)ln(x))/(x^2 +z^2 ))dx∣_(z≠0) A=∫_(−∞) ^0 ((cos(ax)ln(x))/(x^2 +z^2 ))+∫_0 ^∞ ((cos(ax)ln(x))/(x^2 +z^2 ))=Re(2iπ.(e^(−az) /(2iz))ln(iz)) =Reπ(e^(−az) /z)(ln(z)+((iπ)/2))=π(e^(−az) /z)ln(z) ln(−x)=ln(x)+iπ,∀x>0 A2∫_0 ^( ana∞d) ((cos(ax)ln(x))/(x^2 +z^2 ))dx+iπ∫_0 ^∞ ((cos(ax))/(x^2 +z^2 ))xgood ∫_0 ^∞ ((cos(ax)ln(x^2 ))/(x^2 +z^2 ))=f(0)=(π/z)e^(−az) ln(z),∀z,Re(z)>0 C=(π/z)e^(−az) ln(z)−(π/(2z))e^(−az) E_i (az)+(π/(2z))e^(az) E_i (−az) E_i (x)=γ+ln(x)+o(x),x→1 too bee continued not?many times now i think its a good path E_i ...related to chi .. and Shi
$$\left(\mathrm{1}\right) \\ $$$${let}\:{f}\left({b}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({ax}\right){ln}\left({x}^{\mathrm{2}} +{bz}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} },{b}\geqslant\mathrm{0} \\ $$$${f}'\left({b}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{z}^{\mathrm{2}} {cos}\left({ax}\right)}{\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{bz}^{\mathrm{2}} \right)}{dx} \\ $$$$={z}^{\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({ax}\right)}{\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{bz}^{\mathrm{2}} \right)}{dx}=\frac{{z}^{\mathrm{2}} }{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{cos}\left({ax}\right)}{\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{bz}^{\mathrm{2}} \right)}{dx} \\ $$$${A}=\frac{{z}^{\mathrm{2}} }{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{e}^{{iax}} }{\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{bz}^{\mathrm{2}} \right)}{dx}= \\ $$$$={z}^{\mathrm{2}} .{i}\pi.\frac{{e}^{−{az}} }{\mathrm{2}{iz}}.\frac{\mathrm{1}}{\left(\mathrm{1}−{b}\right){z}^{\mathrm{2}} }+{i}\pi.\frac{{e}^{−{az}\sqrt{{b}}} }{{z}^{\mathrm{2}} \left(\mathrm{1}−{b}\right).\left(\mathrm{2}{iz}\sqrt{{b}}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}{z}}\left(\frac{{e}^{−{az}} }{\left(\mathrm{1}−{b}\right)}+\frac{{e}^{−{az}\sqrt{{b}}} }{\left(\mathrm{1}−{b}\right)\sqrt{{b}}}\right) \\ $$$${f}\left({b}\right)=\frac{\pi}{\mathrm{2}{z}}\int\frac{{e}^{−{az}} }{\mathrm{1}−{b}}+\frac{{e}^{−{az}\sqrt{{b}}} }{\left(\mathrm{1}−{b}\right)\sqrt{{b}}}{db} \\ $$$$=\frac{\pi}{\mathrm{2}{z}}\left(−{ln}\left(\mathrm{1}−{b}\right){e}^{−{az}} +\int\frac{{e}^{−{az}\sqrt{{b}}} }{\left(\mathrm{1}−{b}\right)\sqrt{{b}}}{db}\mid_{\sqrt{{b}}={u}_{} } ^{} \right. \\ $$$$\int\frac{{e}^{−{auz}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}.\mathrm{2}{du} \\ $$$$=\int\frac{{e}^{−{auz}} }{\mathrm{1}−{u}}+\frac{{e}^{−{auz}} }{\mathrm{1}+{u}}{du} \\ $$$$=−\int\frac{{e}^{−{az}\left(\mathrm{1}−{x}\right)} }{{x}}{dx}+\int\frac{{e}^{−{az}\left({y}−\mathrm{1}\right)} }{{y}}{dy} \\ $$$$={e}^{−{az}} .−\int\frac{{e}^{{azx}} }{{azx}}{dazx}+{e}^{{az}} \int\frac{{e}^{−{azy}} }{{azy}}{d}\left({azy}\right) \\ $$$$={e}^{−{az}} −{E}_{\mathrm{1}} \left(−{azx}\right)+{e}^{{az}} .{E}_{\mathrm{1}} \left({azy}\right) \\ $$$$=−{e}^{−{az}} {E}_{\mathrm{1}} \left(−{az}\left(\mathrm{1}−\sqrt{{b}}\right)\right)+{e}^{{az}} \left({az}\left(\mathrm{1}+\sqrt{{b}}\right)\right) \\ $$$$−{e}^{{az}} .{E}_{{i}} \left(−{az}\left(\mathrm{1}+\sqrt{{b}}\right)+{e}^{−{az}} {E}_{{i}} \left({az}\left(\mathrm{1}−\sqrt{{b}}\right)\right)\right. \\ $$$${we}\:{get}\: \\ $$$$\frac{\pi}{\mathrm{2}{z}}\left({e}^{−{az}} \left({E}_{{i}} \left({az}\left(\mathrm{1}−\sqrt{{b}}\right)−{ln}\left(\mathrm{1}−{b}\right)\right)−{e}^{{az}} {E}_{{i}} \left(−{az}\left(\mathrm{1}+\sqrt{{b}}\right)\right)\right)+{C}\right. \\ $$$${f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({ax}\right){ln}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({ax}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dx}\mid_{{z}\neq\mathrm{0}} \\ $$$${A}=\int_{−\infty} ^{\mathrm{0}} \frac{{cos}\left({ax}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }+\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({ax}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }={Re}\left(\mathrm{2}{i}\pi.\frac{{e}^{−{az}} }{\mathrm{2}{iz}}{ln}\left({iz}\right)\right) \\ $$$$={Re}\pi\frac{{e}^{−{az}} }{{z}}\left({ln}\left({z}\right)+\frac{{i}\pi}{\mathrm{2}}\right)=\pi\frac{{e}^{−{az}} }{{z}}{ln}\left({z}\right) \\ $$$${ln}\left(−{x}\right)={ln}\left({x}\right)+{i}\pi,\forall{x}>\mathrm{0} \\ $$$${A}\mathrm{2}\int_{\mathrm{0}} ^{\:{ana}\infty{d}} \frac{{cos}\left({ax}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dx}+{i}\pi\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({ax}\right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }{xgood} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({ax}\right){ln}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }={f}\left(\mathrm{0}\right)=\frac{\pi}{{z}}{e}^{−{az}} {ln}\left({z}\right),\forall{z},{Re}\left({z}\right)>\mathrm{0} \\ $$$${C}=\frac{\pi}{{z}}{e}^{−{az}} {ln}\left({z}\right)−\frac{\pi}{\mathrm{2}{z}}{e}^{−{az}} {E}_{{i}} \left({az}\right)+\frac{\pi}{\mathrm{2}{z}}{e}^{{az}} {E}_{{i}} \left(−{az}\right) \\ $$$${E}_{{i}} \left({x}\right)=\gamma+{ln}\left({x}\right)+{o}\left({x}\right),{x}\rightarrow\mathrm{1}\:{too}\:{bee}\:{continued} \\ $$$${not}?{many}\:{times}\:{now} \\ $$$${i}\:{think}\:{its}\:{a}\:{good}\:{path}\:{E}_{{i}} …{related}\:{to}\:{chi}\:.. \\ $$$${and}\:{Shi} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 19/Nov/21
Belle de^� monstration, monsieur!
$$\mathrm{Belle}\:\mathrm{d}\acute {\mathrm{e}monstration},\:\mathrm{monsieur}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *