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 139092 by mathdanisur last updated on 22/Apr/21

Ω=∫_0 ^∞ ((xsinh(πx)e^(−x^2 ) )/(1+x^2 ))dx

$$\Omega=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{xsinh}\left(\pi{x}\right){e}^{−{x}^{\mathrm{2}} } }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Answered by mathmax by abdo last updated on 23/Apr/21

i give this soution but not sure!  Φ=∫_0 ^∞  ((xsh(πx)e^(−x^2 ) )/(x^2  +1))dx ⇒Φ=(1/2)∫_(−∞) ^(+∞)  ((xsh(πx))/(x^2 +1))e^(−x^2 ) dx  w(z)=((zsh(πz)e^(−z^2 ) )/(z^2  +1)) ⇒w(z)=((zsh(πz)e^(−z^2 ) )/((z−i)(z+i)))  ∫_(−∞) ^(+∞)  w(z)dz =2iπRes(w,i)[=2iπ×((ish(πi)e^(−(−1)) )/(2i))=iπsh(πi)e  but  sh(πi)=((e^(i(πi)) −e^(−i(πi)) )/(2i))=((e^(−π) −e^π )/(2i)) ⇒  ∫_(−∞) ^(+∞)  w(z)dz =iπ(((e^(−π) −e^π )/(2i)))e =π(e^(1−π) −e^(1+π) ) ⇒  Φ=(π/2)(e^(1−π) −e^(1+π) )

$$\mathrm{i}\:\mathrm{give}\:\mathrm{this}\:\mathrm{soution}\:\mathrm{but}\:\mathrm{not}\:\mathrm{sure}! \\ $$$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{xsh}\left(\pi\mathrm{x}\right)\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\Phi=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xsh}\left(\pi\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{zsh}\left(\pi\mathrm{z}\right)\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{zsh}\left(\pi\mathrm{z}\right)\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\mathrm{Res}\left(\mathrm{w},\mathrm{i}\right)\left[=\mathrm{2i}\pi×\frac{\mathrm{ish}\left(\pi\mathrm{i}\right)\mathrm{e}^{−\left(−\mathrm{1}\right)} }{\mathrm{2i}}=\mathrm{i}\pi\mathrm{sh}\left(\pi\mathrm{i}\right)\mathrm{e}\right. \\ $$$$\mathrm{but}\:\:\mathrm{sh}\left(\pi\mathrm{i}\right)=\frac{\mathrm{e}^{\mathrm{i}\left(\pi\mathrm{i}\right)} −\mathrm{e}^{−\mathrm{i}\left(\pi\mathrm{i}\right)} }{\mathrm{2i}}=\frac{\mathrm{e}^{−\pi} −\mathrm{e}^{\pi} }{\mathrm{2i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{i}\pi\left(\frac{\mathrm{e}^{−\pi} −\mathrm{e}^{\pi} }{\mathrm{2i}}\right)\mathrm{e}\:=\pi\left(\mathrm{e}^{\mathrm{1}−\pi} −\mathrm{e}^{\mathrm{1}+\pi} \right)\:\Rightarrow \\ $$$$\Phi=\frac{\pi}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{1}−\pi} −\mathrm{e}^{\mathrm{1}+\pi} \right) \\ $$

Commented by mathdanisur last updated on 23/Apr/21

thaks thanks Sir

$${thaks}\:{thanks}\:{Sir} \\ $$

Commented by mathdanisur last updated on 23/Apr/21

Sir, sinh(iπ)=0 ?

$${Sir},\:{sinh}\left({i}\pi\right)=\mathrm{0}\:? \\ $$

Commented by mathmax by abdo last updated on 24/Apr/21

no sir its not like sinx...!

$$\mathrm{no}\:\mathrm{sir}\:\mathrm{its}\:\mathrm{not}\:\mathrm{like}\:\mathrm{sinx}...! \\ $$

Commented by mathdanisur last updated on 24/Apr/21

Sir, omega must be >0.  Your answer is a negative number

$${Sir},\:{omega}\:{must}\:{be}\:>\mathrm{0}. \\ $$$${Your}\:{answer}\:{is}\:{a}\:{negative}\:{number} \\ $$

Commented by mathdanisur last updated on 24/Apr/21

Sir, it′s not like sin, sin(ix)/i=sinh?

$${Sir},\:{it}'{s}\:{not}\:{like}\:{sin},\:{sin}\left({ix}\right)/{i}={sinh}? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com