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Question Number 131529 by mnjuly1970 last updated on 05/Feb/21

...advanced ∗∗∗∗∗∗∗∗∗∗ calculus... prove that ::: :: 𝛗=∫_0 ^( ∞) ((sin(x^4 )ln(x))/x)dx=−((𝛑𝛄)/(32)) note : ∫_0 ^( ∞) ((sin(x)ln(x))/x)dx=^(why???) ((−𝛑𝛄)/2) 𝛗=^(⟨x^4 =t⟩) (1/4)∫_0 ^( ∞) ((sin(t)ln(t^(1/4) ))/t^(1/4) ) ∗(1/t^(3/4) )dt =(1/(16))∫_0 ^( ∞) ((sin(t)ln(t))/t)dt =^(note) ((−𝛑𝛄)/(32)) ( 𝛗=−((𝛑𝛄)/(32)) ) notice:: you will prove that::∫_0 ^( ∞) ((sin(x)ln(x))/x)dx=((−𝛑𝛄)/2) Hint:: ∫_0 ^( ∞) ((sin(x))/(x^( 𝛍) ))dx=^(???) (𝛑/(2𝚪(𝛍)sin(((𝛍𝛑)/2)))) ... m.n.july.1970 ...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{advanced}\:\:\:\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\:\:\:\:{calculus}... \\ $$$$\:\:\:{prove}\:\:{that}\::::\::: \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}^{\mathrm{4}} \right){ln}\left({x}\right)}{{x}}{dx}=−\frac{\boldsymbol{\pi\gamma}}{\mathrm{32}} \\ $$$$\:\:\:\:{note}\::\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx}\overset{{why}???} {=}\:\frac{−\boldsymbol{\pi\gamma}}{\mathrm{2}} \\ $$$$\:\:\:\:\boldsymbol{\phi}\overset{\langle{x}^{\mathrm{4}} ={t}\rangle} {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right){ln}\left({t}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)}{{t}^{\frac{\mathrm{1}}{\mathrm{4}}} }\:\ast\frac{\mathrm{1}}{{t}^{\frac{\mathrm{3}}{\mathrm{4}}} }{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\:\:\infty} \frac{{sin}\left({t}\right){ln}\left({t}\right)}{{t}}{dt}\:\overset{{note}} {=}\frac{−\boldsymbol{\pi\gamma}}{\mathrm{32}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:\:\boldsymbol{\phi}=−\frac{\boldsymbol{\pi\gamma}}{\mathrm{32}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{notice}::\:{you}\:{will}\:{prove}\:{that}::\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx}=\frac{−\boldsymbol{\pi\gamma}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{Hint}::\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right)}{{x}^{\:\boldsymbol{\mu}} \:}{dx}\overset{???} {=}\frac{\boldsymbol{\pi}}{\mathrm{2}\boldsymbol{\Gamma}\left(\boldsymbol{\mu}\right){sin}\left(\frac{\boldsymbol{\mu\pi}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\:{m}.{n}.{july}.\mathrm{1970}\:... \\ $$

Answered by Dwaipayan Shikari last updated on 06/Feb/21

∫_0 ^∞ ((sinx)/x^μ )dx=(1/(Γ(μ)))∫_0 ^∞ ∫_0 ^∞ t^(μ−1) e^(−xt) sinxdxdt =(1/(2iΓ(μ)))∫_0 ^∞ ∫_0 ^∞ t^(μ−1) e^(−x(t−i)) −t^(μ−1) e^(−x(t+i)) dxdt =(1/(2iΓ(μ)))∫_0 ^∞ (t^(μ−1) /(t−i))−(t^(μ−1) /(t+i))dt=(1/(Γ(μ)))∫_0 ^∞ (t^(μ−1) /(t^2 +1))dt =(1/(2Γ(μ)))∫_0 ^∞ (j^((μ/2)−1) /((1+u)))dj=(1/(2Γ(μ))).(π/(sin(((μπ)/2)))) I(μ)=(π/(2Γ(μ)sin(((μπ)/2)))) ⇒I′(μ)=−(π^2 /(4Γ(μ)))cosec(((μπ)/2))cot(((μπ)/2))−(π/(2sin(((μπ)/2)))).((Γ′(μ))/(Γ^2 (μ))) I′(μ)=∫_0 ^∞ ((sinx)/x^μ )log(x)dx I′(1)=∫_0 ^∞ ((sinx)/x)log(x)dx=−(π/2).Γ′(1)=−((πγ)/2)

$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mu} }{dx}=\frac{\mathrm{1}}{\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{\mu−\mathrm{1}} {e}^{−{xt}} {sinxdxdt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{\mu−\mathrm{1}} {e}^{−{x}\left({t}−{i}\right)} −{t}^{\mu−\mathrm{1}} {e}^{−{x}\left({t}+{i}\right)} {dxdt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mu−\mathrm{1}} }{{t}−{i}}−\frac{{t}^{\mu−\mathrm{1}} }{{t}+{i}}{dt}=\frac{\mathrm{1}}{\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mu−\mathrm{1}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \frac{{j}^{\frac{\mu}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)}{dj}=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left(\mu\right)}.\frac{\pi}{{sin}\left(\frac{\mu\pi}{\mathrm{2}}\right)} \\ $$$${I}\left(\mu\right)=\frac{\pi}{\mathrm{2}\Gamma\left(\mu\right){sin}\left(\frac{\mu\pi}{\mathrm{2}}\right)}\:\Rightarrow{I}'\left(\mu\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}\Gamma\left(\mu\right)}{cosec}\left(\frac{\mu\pi}{\mathrm{2}}\right){cot}\left(\frac{\mu\pi}{\mathrm{2}}\right)−\frac{\pi}{\mathrm{2}{sin}\left(\frac{\mu\pi}{\mathrm{2}}\right)}.\frac{\Gamma'\left(\mu\right)}{\Gamma^{\mathrm{2}} \left(\mu\right)} \\ $$$${I}'\left(\mu\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mu} }{log}\left({x}\right){dx} \\ $$$${I}'\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{log}\left({x}\right){dx}=−\frac{\pi}{\mathrm{2}}.\Gamma'\left(\mathrm{1}\right)=−\frac{\pi\gamma}{\mathrm{2}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 05/Feb/21

taydballah mr payan..

$${taydballah}\:{mr}\:{payan}.. \\ $$

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