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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?

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

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