Integration Questions

Question Number 101011 by Dwaipayan Shikari last updated on 29/Jun/20

$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx} \\$$

Commented byRohit@Thakur last updated on 29/Jun/20

$${Use}\:{Laplace}\:{Transform} \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\$$

Commented byRohit@Thakur last updated on 29/Jun/20

$${We}\:{can}\:{evaluate}\:{it}\:{using}\:{Laplace}\:{transform} \\$$ $${L}\left\{{f}\left({t}\right)\right\}=\:{F}\left({s}\right)\:=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{st}} \:{f}\left({t}\right)\:{dt} \\$$ $$\therefore{L}\left\{\frac{{Sint}}{{t}}\right\}\:=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{st}\:} \frac{{Sint}}{{t}}{dt}\:=\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{s}}\right) \\$$ $${F}\left(\mathrm{0}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{Sintdt}}{{t}}\:=\:{lim}\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{s}}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{s}\rightarrow\mathrm{0} \\$$ $$\therefore\int_{\mathrm{0}} ^{\infty} \frac{{Sint}\:{dt}}{{t}}\:=\:\frac{\pi}{\mathrm{2}} \\$$

Commented byDwaipayan Shikari last updated on 29/Jun/20

$${Is}\:{there}\:{any}\:{other}\:{way}? \\$$

Commented bysmridha last updated on 29/Jun/20

$$\boldsymbol{{let}}'{s}\:\boldsymbol{{do}}\:\boldsymbol{{it}}\:\boldsymbol{{by}}\:\boldsymbol{{Feynman}}\:\boldsymbol{{Technique}} \\$$ $$\boldsymbol{{let}}\:\boldsymbol{{I}}\left(\boldsymbol{{a}}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}.\boldsymbol{{e}}^{−\boldsymbol{{ax}}} \boldsymbol{{dx}} \\$$ $$\boldsymbol{{so}}\:\boldsymbol{{I}}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}\:\boldsymbol{{dx}}\:\boldsymbol{{and}}\:\boldsymbol{{I}}\left(\infty\right)=\mathrm{0}\left[\boldsymbol{{as}}\underset{\boldsymbol{{a}}\rightarrow\infty} {\mathrm{lim}}\boldsymbol{{e}}^{−\boldsymbol{{ax}}} =\mathrm{0}\right] \\$$ $$\boldsymbol{{now}}\:\boldsymbol{{diff}}:\boldsymbol{{both}}\:\boldsymbol{{sides}}\:\boldsymbol{{wrt}}\:\boldsymbol{{a}}\:\boldsymbol{{we}}\:\boldsymbol{{get}} \\$$ $$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\left[\boldsymbol{{I}}\left(\boldsymbol{{a}}\right)\right]=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}.\frac{\partial}{\partial\boldsymbol{{a}}}\left(\boldsymbol{{e}}^{−\boldsymbol{{ax}}} \right)\boldsymbol{{dx}} \\$$ $$\boldsymbol{{I}}^{'} \left(\boldsymbol{{a}}\right)=−\int_{\mathrm{0}} ^{\infty} \boldsymbol{{sin}}\left(\boldsymbol{{x}}\right).\boldsymbol{{e}}^{−\boldsymbol{{ax}}} \boldsymbol{{dx}}=−\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{1}} \\$$ $$\boldsymbol{{now}}\:\boldsymbol{{integrating}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\:\boldsymbol{{wrt}}\:\boldsymbol{{a}} \\$$ $$\int_{\mathrm{0}} ^{\infty} \boldsymbol{{I}}^{'} \left(\boldsymbol{{a}}\right)\boldsymbol{{da}}=−\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{da}}}{\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{1}}=−\left[\mathrm{tan}^{−\mathrm{1}} \left(\boldsymbol{{a}}\right)\right]_{\mathrm{0}} ^{\infty} =−\frac{\boldsymbol{\pi}}{\mathrm{2}} \\$$ $$\boldsymbol{{I}}\left(\infty\right)−\boldsymbol{{I}}\left(\mathrm{0}\right)=−\frac{\boldsymbol{\pi}}{\mathrm{2}} \\$$ $$\mathrm{0}−\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}}=−\frac{\boldsymbol{\pi}}{\mathrm{2}} \\$$ $$\boldsymbol{{so}}\:\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}. \\$$ $$\\$$

Commented bysmridha last updated on 29/Jun/20

$$\boldsymbol{{another}}\:\boldsymbol{{way}} \\$$ $$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\boldsymbol{{dx}} \\$$ $$\boldsymbol{{by}}\:\boldsymbol{{Cauchy}}\:\boldsymbol{{principal}}\:\boldsymbol{{value}}\:\boldsymbol{{theorem}} \\$$ $$\boldsymbol{{we}}\:\boldsymbol{{get}} \\$$ $$\boldsymbol{{I}}=\boldsymbol{{p}}\int_{−\infty} ^{+\infty} \frac{\boldsymbol{{e}}^{\boldsymbol{{ix}}} }{\mathrm{2}\boldsymbol{{ix}}}\boldsymbol{{dx}} \\$$ $$\boldsymbol{{now}}\:\boldsymbol{{two}}\:\boldsymbol{{semi}}−\boldsymbol{{infinite}}\:\boldsymbol{{segments}} \\$$ $$\boldsymbol{{costituting}}\:\boldsymbol{{the}}\:\boldsymbol{{pricipal}}\:\boldsymbol{{value}}\:\boldsymbol{{integral}} \\$$ $$\left(\boldsymbol{{i}}\right)\boldsymbol{{semicircle}}\:\boldsymbol{{C}}_{\boldsymbol{{R}}} \boldsymbol{{of}}\:\boldsymbol{{redius}}\left(\boldsymbol{{R}}\Rightarrow\infty\right) \\$$ $$\left(\boldsymbol{{ii}}\right)\boldsymbol{{semicircle}}\:\boldsymbol{{C}}_{\boldsymbol{{r}}} \:\boldsymbol{{of}}\:\boldsymbol{{redius}}\:\left(\boldsymbol{{r}}\Rightarrow\mathrm{0}\right) \\$$ $$\:\boldsymbol{{is}}\:\boldsymbol{{taken}}\:\boldsymbol{{clockwise}}\:\boldsymbol{{now}} \\$$ $$\oint\frac{\boldsymbol{{e}}^{\boldsymbol{{iz}}} }{\mathrm{2}\boldsymbol{{iz}}}\boldsymbol{{dz}}=\boldsymbol{{I}}+\int_{\boldsymbol{{c}}_{\boldsymbol{{r}}} } \frac{\boldsymbol{{e}}^{\boldsymbol{{iz}}} }{\mathrm{2}\boldsymbol{{iz}}}\boldsymbol{{dz}}+\int_{\boldsymbol{{c}}_{\boldsymbol{{R}}} } \frac{\boldsymbol{{e}}^{\boldsymbol{{iz}}} }{\mathrm{2}\boldsymbol{{iz}}}\boldsymbol{{dz}}=\mathrm{0} \\$$ $$\boldsymbol{{by}}\:\boldsymbol{{J}}{o}\boldsymbol{{rdan}}'{s}\:\boldsymbol{{lemma}}\:\boldsymbol{{the}}\:\boldsymbol{{integral}} \\$$ $$\boldsymbol{{over}}\:\boldsymbol{{C}}_{{R}} \:\boldsymbol{{vanishes}}. \\$$ $$\boldsymbol{{the}}\:\boldsymbol{{clockwise}}\:\boldsymbol{{path}}\:\boldsymbol{{C}}_{\boldsymbol{{r}}} \:\boldsymbol{{half}}\:\boldsymbol{{way}} \\$$ $$\boldsymbol{{around}}\:\boldsymbol{{the}}\:\boldsymbol{{pole}}\:\boldsymbol{{z}}=\mathrm{0}\:\boldsymbol{{contributes}} \\$$ $$\boldsymbol{{half}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{full}}\:\boldsymbol{{circuit}} \\$$ $$\boldsymbol{{so}}\:\:\boldsymbol{{the}}\:\boldsymbol{{residue}}\:\boldsymbol{{around}}\:\boldsymbol{{the}}\:\boldsymbol{{pole}} \\$$ $$\boldsymbol{{z}}=\mathrm{0}\:\boldsymbol{{is}}\:\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{i}}}\:\boldsymbol{{so}}\:\int_{\boldsymbol{{c}}_{\boldsymbol{{r}}} } =−\boldsymbol{\pi{i}}.\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{i}}}=−\frac{\boldsymbol{\pi}}{\mathrm{2}}\: \\$$ $$\boldsymbol{{so}}\:\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}. \\$$