Question Number 58791 by Tawa1 last updated on 30/Apr/19
$$\:\boldsymbol{\mathrm{Show}}\:\boldsymbol{\mathrm{that}}:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\infty} \:\:\:\frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\:\:\:=\:\:\frac{\pi}{\mathrm{2}} \\ $$
Answered by tanmay last updated on 30/Apr/19
$${it}\:{has}\:{some}\:{trick}\:{to}\:{solve}… \\ $$$${F}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{ax}} {sinx}}{{x}}{dx} \\ $$$$\frac{{dF}}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×\frac{\partial}{\partial{a}}\left({e}^{−{ax}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×{e}^{−{ax}} ×−{xdx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} −{sinx}×{e}^{−{ax}} {dx} \\ $$$${now}\:{let} \\ $$$${p}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{cosxdx} \\ $$$${q}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{sinxdx} \\ $$$${p}+{iq}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{e}^{{ix}} {dx} \\ $$$${p}+{iq}=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}\left({a}−{i}\right)} {dx} \\ $$$$=\mid\frac{{e}^{−{x}\left({a}−{i}\right)} }{−\left({a}−{i}\right)}\mid_{\mathrm{0}} ^{\infty} =\frac{−\mathrm{1}}{−{a}+{i}}=\frac{\mathrm{1}}{{a}−{i}}=\frac{{a}+{i}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$${p}=\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\:{and}\:\:{iq}={i}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${q}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} −{e}^{−{ax}} ×{sinxdx}=−{q}=\frac{−\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} −{e}^{−{ax}} ×{sinxdx}=\frac{−\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}=\frac{{dF}\left({a}\right)}{{da}} \\ $$$$\int{dF}\left({a}\right)=\left(−\mathrm{1}\right)\int\frac{{da}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$${F}\left({a}\right)=\left(−\mathrm{1}\right){tan}^{−\mathrm{1}} {a}+{c} \\ $$$$ \\ $$$${as}\:{a}\rightarrow\infty\:{F}\left({a}\right)\rightarrow\mathrm{0} \\ $$$$\mathrm{0}=\left(−\mathrm{1}\right){tan}^{−\mathrm{1}} \left(\infty\right)+{c} \\ $$$$\mathrm{0}=−\frac{\pi}{\mathrm{2}}+{c} \\ $$$${c}=\frac{\pi}{\mathrm{2}} \\ $$$${F}\left({a}\right)=−{tan}^{−\mathrm{1}} \left({a}\right)+\frac{\pi}{\mathrm{2}} \\ $$$${nowF}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×\frac{{sinx}}{{x}}{dx}=−{tan}^{−\mathrm{1}} \left({a}\right)+\frac{\pi}{\mathrm{2}} \\ $$$${now}\:{if}\:{you}\:{put}\:{a}=\mathrm{0} \\ $$$${we}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=−{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)+\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 30/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay last updated on 30/Apr/19
$${than}\:{you}…{blessing}\:{shower}\:{to}\:{all} \\ $$