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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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