Question Number 116436 by mnjuly1970 last updated on 04/Oct/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:{evaluate}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\:\int_{\:\mathrm{0}} ^{\:\:\infty} \:\:\left(\frac{{sin}\left({x}\right).{sin}\left(\mathrm{2}{x}\right)}{{x}}\right)\:{dx}\:=???\: \\ $$$$\:\:\:\:\:\:\:\:\:…\:{m}.{n}.\mathrm{1970}… \\ $$$$\: \\ $$$$ \\ $$
Commented by Bird last updated on 04/Oct/20
$${it}\:{seems}\:{that}\:{integral}\:{is}\:{divergent}! \\ $$
Commented by mathdave last updated on 04/Oct/20
$${is}\:{not}\:{divergent}\:{oooo}\: \\ $$
Answered by mnjuly1970 last updated on 04/Oct/20
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Answered by Bird last updated on 05/Oct/20
![let take a try we have sinx.sin(2x)=cos((π/2)−x)cos((π/2)−2x) =(1/2){cos(π−3x)+cos(x)} =(1/2){cosx−cos(3x)} ⇒ I =(1/2)∫_0 ^∞ ((cosx−cos(3x))/x)dx I =∫_0 ^∞ ((1−cos(3x))/(2x))dx−∫_0 ^∞ ((1−cosx)/(2x))dx =(1/2)(u−v) let f(a) =∫_0 ^∞ ((1−cosx)/x)e^(−ax) dx with a>0 f^′ (a) =−∫_0 ^∞ (1−cosx)e^(−ax) dx =∫_0 ^∞ e^(−ax) cosxdx−∫_0 ^∞ e^(−ax) dx ∫_0 ^∞ e^(−ax) dx=[−(1/a)e^(−ax) ]_0 ^∞ =(1/a) ∫_0 ^∞ e^(−ax+ix) dx =∫_0 ^∞ e^((−a+i)x) dx =[(1/(−a+i)) e^((−a+i)x) ]_0 ^∞ =−(1/(−a+i))=(1/(a−i)) =((a+i)/(a^2 +1)) ⇒∫_0 ^∞ e^(−ax) cosxdx =Re(...) =(a/(1+a^2 )) ⇒ f^′ (a) =(a/(1+a^2 ))−(1/a) ⇒ f(a) =(1/2)ln(1+a^2 )−lna +c =ln(((√(1+a^2 ))/a)) +c lim_(a→+∞) f(a) =0 =c ⇒ f(a) =ln(((√(1+a^2 ))/a)) ∫_0 ^∞ ((1−cosx)/x)dx =lim_(a→0) f(a) dont exist also ∫_0 ^∞ ((1−cos(3x))/x)dxdont exist but we see ∫_0 ^∞ ((1−cos(3x))/x)dx=_(3x=t) 3∫_0 ^∞ ((1−cost)/t)(dt/3) =∫_0 ^∞ ((1−cosx)/x)dx I =0](https://www.tinkutara.com/question/Q116583.png)
$${let}\:{take}\:{a}\:{try}\:\:{we}\:{have} \\ $$$${sinx}.{sin}\left(\mathrm{2}{x}\right)={cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right){cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\pi−\mathrm{3}{x}\right)+{cos}\left({x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cosx}−{cos}\left(\mathrm{3}{x}\right)\right\}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{cosx}−{cos}\left(\mathrm{3}{x}\right)}{{x}}{dx} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)}{\mathrm{2}{x}}{dx}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{cosx}}{\mathrm{2}{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({u}−{v}\right) \\ $$$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cosx}}{{x}}{e}^{−{ax}} {dx}\:{with}\:{a}>\mathrm{0} \\ $$$${f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\left(\mathrm{1}−{cosx}\right){e}^{−{ax}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ax}} {cosxdx}−\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ax}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ax}} {dx}=\left[−\frac{\mathrm{1}}{{a}}{e}^{−{ax}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{{a}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ax}+{ix}} {dx}\:=\int_{\mathrm{0}} ^{\infty} {e}^{\left(−{a}+{i}\right){x}} {dx} \\ $$$$=\left[\frac{\mathrm{1}}{−{a}+{i}}\:{e}^{\left(−{a}+{i}\right){x}} \right]_{\mathrm{0}} ^{\infty} =−\frac{\mathrm{1}}{−{a}+{i}}=\frac{\mathrm{1}}{{a}−{i}} \\ $$$$=\frac{{a}+{i}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} \:{cosxdx} \\ $$$$={Re}\left(…\right)\:=\frac{{a}}{\mathrm{1}+{a}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\frac{{a}}{\mathrm{1}+{a}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)−{lna}\:+{c} \\ $$$$={ln}\left(\frac{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}\right)\:+{c} \\ $$$${lim}_{{a}\rightarrow+\infty} {f}\left({a}\right)\:=\mathrm{0}\:={c}\:\Rightarrow \\ $$$${f}\left({a}\right)\:={ln}\left(\frac{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cosx}}{{x}}{dx}\:={lim}_{{a}\rightarrow\mathrm{0}} \:{f}\left({a}\right) \\ $$$${dont}\:{exist}\:{also}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)}{{x}}{dxdont} \\ $$$${exist}\:{but}\:{we}\:{see} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)}{{x}}{dx}=_{\mathrm{3}{x}={t}} \:\mathrm{3}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cost}}{{t}}\frac{{dt}}{\mathrm{3}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cosx}}{{x}}{dx}\:{I}\:=\mathrm{0} \\ $$
Commented by Bird last updated on 05/Oct/20
$${u}={v}\:\Rightarrow{I}\:=\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 05/Oct/20
![solution: method 1 : I =L [((sin(x).sin(2x)/x)]_(s=0) = L [sin(x) sin(2x)]=(1/2) L [cos(x−2x) −cos(x+2x)] =(1/2) L [cos(x)]− (1/2) L [cos(3x)] =(1/2)∗ (s/(s^2 +1)) −(1/2)∗ (s/(s^2 +9)) I = {(1/2)∫_s ^( ∞) (u/(u^2 +1)) du − (1/2) ∫_s ^( ∞) (u/(u^2 +9))du}_(s=0) ={(1/4) [ln (((u^2 +1)/(u^2 +9)))]_( u=s) ^( u=∞) = (1/4) ln((1/9))−(1/4) ln(((s^2 +1)/(s^2 +9)) )}_(s=0) =[((−1)/2) ln(3) − (1/4)ln(((s^2 +1)/(s^2 +9)))]_( s=0) = thank you your effort and work is admirable ...](https://www.tinkutara.com/question/Q116624.png)
$${solution}:\: \\ $$$${method}\:\mathrm{1}\::\:\mathrm{I}\:=\mathscr{L}\:\left[\frac{{sin}\left({x}\right).{sin}\left(\mathrm{2}{x}\right.}{{x}}\right]_{{s}=\mathrm{0}} \: \\ $$$$\:\:\:=\:\mathscr{L}\:\left[{sin}\left({x}\right)\:{sin}\left(\mathrm{2}{x}\right)\right]=\frac{\mathrm{1}}{\mathrm{2}}\:\mathscr{L}\:\left[{cos}\left({x}−\mathrm{2}{x}\right)\:−{cos}\left({x}+\mathrm{2}{x}\right)\right]\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\mathscr{L}\:\left[{cos}\left({x}\right)\right]−\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathscr{L}\:\left[{cos}\left(\mathrm{3}{x}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\ast\:\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\ast\:\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{9}} \\ $$$$\mathrm{I}\:=\:\left\{\frac{\mathrm{1}}{\mathrm{2}}\int_{{s}} ^{\:\infty} \frac{{u}}{{u}^{\mathrm{2}} +\mathrm{1}}\:{du}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{s}} ^{\:\infty} \frac{{u}}{{u}^{\mathrm{2}} +\mathrm{9}}{du}\right\}_{{s}=\mathrm{0}} \: \\ $$$$=\left\{\frac{\mathrm{1}}{\mathrm{4}}\:\left[{ln}\:\left(\frac{{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{9}}\right)\right]_{\:{u}={s}} ^{\:{u}=\infty} =\:\frac{\mathrm{1}}{\mathrm{4}}\:{ln}\left(\frac{\mathrm{1}}{\mathrm{9}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\:{ln}\left(\frac{{s}^{\mathrm{2}} +\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{9}}\:\right)\right\}_{{s}=\mathrm{0}} \: \\ $$$$\:=\left[\frac{−\mathrm{1}}{\mathrm{2}}\:{ln}\left(\mathrm{3}\right)\:−\:\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{{s}^{\mathrm{2}} +\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{9}}\right)\right]_{\:{s}=\mathrm{0}} \\ $$$$\:= \\ $$$$\:{thank}\:{you}\:{your}\:{effort}\:\:{and} \\ $$$${work}\:\:{is}\:\:{admirable}\:… \\ $$
Answered by mathdave last updated on 05/Oct/20
$${solution} \\ $$$${we}\:{know}\:\mathrm{sin}{x}\mathrm{sin2}{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}{x}−\mathrm{cos3}{x}\right) \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{x}\mathrm{sin2}{x}}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{cos}{x}−\mathrm{cos3}{x}}{{x}}\right){dx} \\ $$$${from}\:{maz}\:{identity}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} {f}\left({t}\right){G}\left({t}\right){dt}=\int^{\infty} {f}\left({s}\right){g}\left({s}\right){ds}\:\:{but}\:{note}\:\: \\ $$$${L}\left({f}\left({t}\right)\right)={f}\left({s}\right)\:\:{and}\:\:{L}^{−\mathrm{1}} \left({G}\left({t}\right)\right)={g}\left({s}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {L}\left(\mathrm{cos}{x}−\mathrm{cos3}{x}\right){L}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)_{{s}={x}} {ds}\:\:\:{but}\:{l}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{1} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{1}}−\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{9}}\right){ds}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{s}}{\mathrm{2}\left({s}^{\mathrm{2}} +\mathrm{1}\right)}{ds}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{s}}{\mathrm{2}\left({s}^{\mathrm{2}} +\mathrm{9}\right)}{ds}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\frac{{s}^{\mathrm{2}} +\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{9}}\right)_{\mathrm{0}} ^{\infty} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{0}−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{9}}\right)\right)=\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{2}} \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{x}\mathrm{sin2}{x}}{{x}}{dx}=\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{2}} \\ $$$${by}\:{mathdave}\left(\mathrm{05}/\mathrm{10}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 06/Oct/20
$${thank}\:{you} \\ $$$${very}\:{nice}\:{as}\:\:{always}… \\ $$