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Question Number 58791 by Tawa1 last updated on 30/Apr/19
 Show that:         ∫_( 0) ^( ∞)    ((sin(x))/x)   =  (π/2)
$$\:\boldsymbol{\mathrm{Show}}\:\boldsymbol{\mathrm{that}}:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\infty} \:\:\:\frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\:\:\:=\:\:\frac{\pi}{\mathrm{2}} \\ $$
Answered by tanmay last updated on 30/Apr/19
it has some trick to solve...  F(a)=∫_0 ^∞ ((e^(−ax) sinx)/x)dx  (dF/da)=∫_0 ^∞ ((sinx)/x)×(∂/∂a)(e^(−ax) )dx  =∫_0 ^∞ ((sinx)/x)×e^(−ax) ×−xdx  =∫_0 ^∞ −sinx×e^(−ax) dx  now let  p=∫_0 ^∞ e^(−ax) ×cosxdx  q=∫_0 ^∞ e^(−ax) ×sinxdx  p+iq=∫_0 ^∞ e^(−ax) ×e^(ix) dx  p+iq=∫_0 ^∞ e^(−x(a−i)) dx  =∣(e^(−x(a−i)) /(−(a−i)))∣_0 ^∞ =((−1)/(−a+i))=(1/(a−i))=((a+i)/(a^2 +1))  p=(a/(a^2 +1))  and  iq=i((1/(a^2 +1)))  q=(1/(a^2 +1))  ∫_0 ^∞ −e^(−ax) ×sinxdx=−q=((−1)/(a^2 +1))  ∫_0 ^∞ −e^(−ax) ×sinxdx=((−1)/(a^2 +1))=((dF(a))/da)  ∫dF(a)=(−1)∫(da/(a^2 +1))  F(a)=(−1)tan^(−1) a+c    as a→∞ F(a)→0  0=(−1)tan^(−1) (∞)+c  0=−(π/2)+c  c=(π/2)  F(a)=−tan^(−1) (a)+(π/2)  nowF(a)=∫_0 ^∞ e^(−ax) ×((sinx)/x)dx=−tan^(−1) (a)+(π/2)  now if you put a=0  we get  ∫_0 ^∞ ((sinx)/x)dx=−tan^(−1) (0)+(π/2)  ∫_0 ^∞ ((sinx)/x)dx=(π/2)
$${it}\:{has}\:{some}\:{trick}\:{to}\:{solve}… \\ $$$${F}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{ax}} {sinx}}{{x}}{dx} \\ $$$$\frac{{dF}}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×\frac{\partial}{\partial{a}}\left({e}^{−{ax}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×{e}^{−{ax}} ×−{xdx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} −{sinx}×{e}^{−{ax}} {dx} \\ $$$${now}\:{let} \\ $$$${p}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{cosxdx} \\ $$$${q}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{sinxdx} \\ $$$${p}+{iq}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{e}^{{ix}} {dx} \\ $$$${p}+{iq}=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}\left({a}−{i}\right)} {dx} \\ $$$$=\mid\frac{{e}^{−{x}\left({a}−{i}\right)} }{−\left({a}−{i}\right)}\mid_{\mathrm{0}} ^{\infty} =\frac{−\mathrm{1}}{−{a}+{i}}=\frac{\mathrm{1}}{{a}−{i}}=\frac{{a}+{i}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$${p}=\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\:{and}\:\:{iq}={i}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${q}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} −{e}^{−{ax}} ×{sinxdx}=−{q}=\frac{−\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} −{e}^{−{ax}} ×{sinxdx}=\frac{−\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}=\frac{{dF}\left({a}\right)}{{da}} \\ $$$$\int{dF}\left({a}\right)=\left(−\mathrm{1}\right)\int\frac{{da}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$${F}\left({a}\right)=\left(−\mathrm{1}\right){tan}^{−\mathrm{1}} {a}+{c} \\ $$$$ \\ $$$${as}\:{a}\rightarrow\infty\:{F}\left({a}\right)\rightarrow\mathrm{0} \\ $$$$\mathrm{0}=\left(−\mathrm{1}\right){tan}^{−\mathrm{1}} \left(\infty\right)+{c} \\ $$$$\mathrm{0}=−\frac{\pi}{\mathrm{2}}+{c} \\ $$$${c}=\frac{\pi}{\mathrm{2}} \\ $$$${F}\left({a}\right)=−{tan}^{−\mathrm{1}} \left({a}\right)+\frac{\pi}{\mathrm{2}} \\ $$$${nowF}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×\frac{{sinx}}{{x}}{dx}=−{tan}^{−\mathrm{1}} \left({a}\right)+\frac{\pi}{\mathrm{2}} \\ $$$${now}\:{if}\:{you}\:{put}\:{a}=\mathrm{0} \\ $$$${we}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=−{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)+\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 30/Apr/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay last updated on 30/Apr/19
than you...blessing shower to all
$${than}\:{you}…{blessing}\:{shower}\:{to}\:{all} \\ $$

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