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a-b-R-e-iax-1-e-ibx-1-x-2-dx-

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Question Number 219428 by Nicholas666 last updated on 24/Apr/25
a , b, ∈ R ∫_( −∞) ^( ∞) (((e^(iax) −1)(e^(ibx) −1))/x^2 ) dx
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:,\:{b},\:\in\:\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{\:−\infty} ^{\:\infty} \frac{\left({e}^{{iax}} −\mathrm{1}\right)\left({e}^{{ibx}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$ \\ $$
Commented by Ghisom last updated on 25/Apr/25
this should be zero (?)
$$\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{zero}\:\left(?\right) \\ $$
Answered by vnm last updated on 25/Apr/25
I_1 (p,q)=∫_(−∞) ^(+∞) ((sinpxcosqx)/x)dx=∫_(−∞) ^(+∞) (((1/2)(sin(px+qx)+sin(px−qx)))/x)dx=(1/2)∫_(−∞) ^(+∞) ((sin(p+q)x)/x)dx+(1/2)∫_(−∞) ^(+∞) ((sin(p−q)x)/x)dx= (π/2)(sgn(p+q)+sgn(p−q)) I_2 (p)=∫_(−∞) ^(+∞) ((sin^2 px)/x^2 )dx=∫_(−∞) ^(+∞) (d/dx)(−(1/x))sin^2 pxdx=0+∫_(−∞) ^(+∞) (1/x)(d/dx)sin^2 pxdx=∫_(−∞) ^(+∞) ((2psinpxcospx)/x)dx=p∫_(−∞) ^(+∞) ((sin2px)/(2px))d(2px)= π∣p∣ I_3 (p,q)=∫_(−∞) ^(+∞) ((sinpxsinqx)/x^2 )dx=∫_(−∞) ^(+∞) (d/dx)(−(1/x))sinpxsinqxdx=0+∫_(−∞) ^(+∞) (1/x)(d/dx)(sinpxsinqx)dx=∫_(−∞) ^(+∞) (1/x)(pcospxsinqx+qsinpxcosqx)dx= p∫_(−∞) ^(+∞) ((sinqxcospx)/x)dx+q∫_(−∞) ^(+∞) ((sinpxcosqx)/x)dx=pI_1 (q,p)+qI_1 (p,q) (π/2)(p(sgn(q+p)+sgn(q−p))+q(sgn(p+q)+sgn(p−q)))= π(p((sgn(p+q)−sgn(p−q))/2)+q((sgn(p+q)+sgn(p−q))/2)) ∫_(−∞) ^(+∞) (((e^(iax) −1)(e^(ibx) −1))/x^2 )dx=∫_(−∞) ^(+∞) (((cosax+isinax−1)(cosbx+isinbx−1))/x^2 )dx= ∫_(−∞) ^(+∞) (((cosax−1)(cosbx−1))/x^2 )dx−I_3 (a,b)=∫_(−∞) ^(+∞) (((−2sin^2 ((ax)/2))(−2sin^2 ((bx)/2)))/x^2 )dx−I_3 (a,b)=4∫_(−∞) ^(+∞) ((sin^2 ((ax)/2)(1−cos^2 ((bx)/2)))/x^2 )dx=4I_2 ((a/2))−4∫_(−∞) ^(+∞) ((sin^2 ((ax)/2)cos^2 ((bx)/2))/x^2 )dx−I_3 (a,b)= 4I_2 ((a/2))−4∫_(−∞) ^(+∞) (((sin((ax)/2)cos((bx)/2))^2 )/x^2 )dx−I_3 (a,b)=4I_2 ((a/2))−4∫_(−∞) ^(+∞) ((((1/2)(sin(((a+b)x)/2)+sin(((a−b)x)/2)))^2 )/x^2 )dx−I_3 (a,b)= 4I_2 ((a/2))−∫_(−∞) ^(+∞) (((sin^2 (((a+b)x)/2))/x^2 )+2((sin(((a+b)x)/2)sin(((a−b)x)/2))/x^2 )+((sin^2 (((a−b)x)/2))/x^2 ))dx−I_3 (a,b)= 4I_2 ((a/2))−I_2 (((a+b)/2))−2I_3 (((a+b)/2),((a−b)/2))−I_2 (((a−b)/2))−I_3 (a,b)
$${I}_{\mathrm{1}} \left({p},{q}\right)=\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}{px}\mathrm{cos}{qx}}{{x}}\mathrm{d}{x}=\int_{−\infty} ^{+\infty} \frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\left({px}+{qx}\right)+\mathrm{sin}\left({px}−{qx}\right)\right)}{{x}}\mathrm{d}{x}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}\left({p}+{q}\right){x}}{{x}}\mathrm{d}{x}+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}\left({p}−{q}\right){x}}{{x}}\mathrm{d}{x}= \\ $$$$\frac{\pi}{\mathrm{2}}\left(\mathrm{sgn}\left({p}+{q}\right)+\mathrm{sgn}\left({p}−{q}\right)\right) \\ $$$${I}_{\mathrm{2}} \left({p}\right)=\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}^{\mathrm{2}} {px}}{{x}^{\mathrm{2}} }\mathrm{d}{x}=\int_{−\infty} ^{+\infty} \frac{\mathrm{d}}{\mathrm{d}{x}}\left(−\frac{\mathrm{1}}{{x}}\right)\mathrm{sin}^{\mathrm{2}} {px}\mathrm{d}{x}=\mathrm{0}+\int_{−\infty} ^{+\infty} \frac{\mathrm{1}}{{x}}\frac{\mathrm{d}}{\mathrm{d}{x}}\mathrm{sin}^{\mathrm{2}} {px}\mathrm{d}{x}=\int_{−\infty} ^{+\infty} \frac{\mathrm{2}{p}\mathrm{sin}{px}\mathrm{cos}{px}}{{x}}\mathrm{d}{x}={p}\int_{−\infty} ^{+\infty} \frac{\mathrm{sin2}{px}}{\mathrm{2}{px}}\mathrm{d}\left(\mathrm{2}{px}\right)= \\ $$$$\pi\mid{p}\mid \\ $$$${I}_{\mathrm{3}} \left({p},{q}\right)=\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}{px}\mathrm{sin}{qx}}{{x}^{\mathrm{2}} }\mathrm{d}{x}=\int_{−\infty} ^{+\infty} \frac{\mathrm{d}}{\mathrm{d}{x}}\left(−\frac{\mathrm{1}}{{x}}\right)\mathrm{sin}{px}\mathrm{sin}{qx}\mathrm{d}{x}=\mathrm{0}+\int_{−\infty} ^{+\infty} \frac{\mathrm{1}}{{x}}\frac{\mathrm{d}}{\mathrm{d}{x}}\left(\mathrm{sin}{px}\mathrm{sin}{qx}\right)\mathrm{d}{x}=\int_{−\infty} ^{+\infty} \frac{\mathrm{1}}{{x}}\left({p}\mathrm{cos}{px}\mathrm{sin}{qx}+{q}\mathrm{sin}{px}\mathrm{cos}{qx}\right)\mathrm{d}{x}= \\ $$$${p}\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}{qx}\mathrm{cos}{px}}{{x}}\mathrm{d}{x}+{q}\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}{px}\mathrm{cos}{qx}}{{x}}\mathrm{d}{x}={pI}_{\mathrm{1}} \left({q},{p}\right)+{qI}_{\mathrm{1}} \left({p},{q}\right) \\ $$$$\frac{\pi}{\mathrm{2}}\left({p}\left(\mathrm{sgn}\left({q}+{p}\right)+\mathrm{sgn}\left({q}−{p}\right)\right)+{q}\left(\mathrm{sgn}\left({p}+{q}\right)+\mathrm{sgn}\left({p}−{q}\right)\right)\right)= \\ $$$$\pi\left({p}\frac{\mathrm{sgn}\left({p}+{q}\right)−\mathrm{sgn}\left({p}−{q}\right)}{\mathrm{2}}+{q}\frac{\mathrm{sgn}\left({p}+{q}\right)+\mathrm{sgn}\left({p}−{q}\right)}{\mathrm{2}}\right) \\ $$$$\int_{−\infty} ^{+\infty} \frac{\left({e}^{{iax}} −\mathrm{1}\right)\left({e}^{{ibx}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} }\mathrm{d}{x}=\int_{−\infty} ^{+\infty} \frac{\left(\mathrm{cos}{ax}+{i}\mathrm{sin}{ax}−\mathrm{1}\right)\left(\mathrm{cos}{bx}+{i}\mathrm{sin}{bx}−\mathrm{1}\right)}{{x}^{\mathrm{2}} }\mathrm{d}{x}= \\ $$$$\int_{−\infty} ^{+\infty} \frac{\left(\mathrm{cos}{ax}−\mathrm{1}\right)\left(\mathrm{cos}{bx}−\mathrm{1}\right)}{{x}^{\mathrm{2}} }\mathrm{d}{x}−{I}_{\mathrm{3}} \left({a},{b}\right)=\int_{−\infty} ^{+\infty} \frac{\left(−\mathrm{2sin}^{\mathrm{2}} \frac{{ax}}{\mathrm{2}}\right)\left(−\mathrm{2sin}^{\mathrm{2}} \frac{{bx}}{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\mathrm{d}{x}−{I}_{\mathrm{3}} \left({a},{b}\right)=\mathrm{4}\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}^{\mathrm{2}} \frac{{ax}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \frac{{bx}}{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\mathrm{d}{x}=\mathrm{4}{I}_{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)−\mathrm{4}\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}^{\mathrm{2}} \frac{{ax}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \frac{{bx}}{\mathrm{2}}}{{x}^{\mathrm{2}} }\mathrm{d}{x}−{I}_{\mathrm{3}} \left({a},{b}\right)= \\ $$$$\mathrm{4}{I}_{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)−\mathrm{4}\int_{−\infty} ^{+\infty} \frac{\left(\mathrm{sin}\frac{{ax}}{\mathrm{2}}\mathrm{cos}\frac{{bx}}{\mathrm{2}}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} }\mathrm{d}{x}−{I}_{\mathrm{3}} \left({a},{b}\right)=\mathrm{4}{I}_{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)−\mathrm{4}\int_{−\infty} ^{+\infty} \frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\frac{\left({a}+{b}\right){x}}{\mathrm{2}}+\mathrm{sin}\frac{\left({a}−{b}\right){x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} }\mathrm{d}{x}−{I}_{\mathrm{3}} \left({a},{b}\right)= \\ $$$$\mathrm{4}{I}_{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)−\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{sin}^{\mathrm{2}} \frac{\left({a}+{b}\right){x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }+\mathrm{2}\frac{\mathrm{sin}\frac{\left({a}+{b}\right){x}}{\mathrm{2}}\mathrm{sin}\frac{\left({a}−{b}\right){x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }+\frac{\mathrm{sin}^{\mathrm{2}} \frac{\left({a}−{b}\right){x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }\right)\mathrm{d}{x}−{I}_{\mathrm{3}} \left({a},{b}\right)= \\ $$$$\mathrm{4}{I}_{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)−{I}_{\mathrm{2}} \left(\frac{{a}+{b}}{\mathrm{2}}\right)−\mathrm{2}{I}_{\mathrm{3}} \left(\frac{{a}+{b}}{\mathrm{2}},\frac{{a}−{b}}{\mathrm{2}}\right)−{I}_{\mathrm{2}} \left(\frac{{a}−{b}}{\mathrm{2}}\right)−{I}_{\mathrm{3}} \left({a},{b}\right) \\ $$
Answered by aleks041103 last updated on 29/Apr/25
f(a,b)=∫_(−∞) ^( ∞) (((e^(iax) −1)(e^(ibx) −1))/x^2 ) ⇒(∂^2 f/(∂a∂b)) = −∫_(−∞) ^( ∞) e^(i(a+b)x) dx = −2πδ(a+b) ⇒(∂f/∂a) = f_1 ′(a)−2πH(a+b) ⇒f(a,b) = f_1 (a)+f_2 (b)−2π(a+b)H(a+b) f(a,b)=f(b,a) (obvious) ⇒f_1 (a)=f_2 (a)+c, c is just a number ⇒f(a,b)=g(a)+g(b)−2π(a+b)H(a+b) f(a,0)=0=g(a)+g(0)−2πaH(a) ⇒g(a)=2πaH(a)−g(0) ⇒2g(0)=0⇒g(a)=2πaH(a) ⇒∫_(−∞) ^( ∞) ... = 2π(aH(a)+bH(b)−(a+b)H(a+b))
$${f}\left({a},{b}\right)=\int_{−\infty} ^{\:\infty} \frac{\left({e}^{{iax}} −\mathrm{1}\right)\left({e}^{{ibx}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\partial^{\mathrm{2}} {f}}{\partial{a}\partial{b}}\:=\:−\int_{−\infty} ^{\:\infty} {e}^{{i}\left({a}+{b}\right){x}} {dx}\:=\:−\mathrm{2}\pi\delta\left({a}+{b}\right) \\ $$$$\Rightarrow\frac{\partial{f}}{\partial{a}}\:=\:{f}_{\mathrm{1}} '\left({a}\right)−\mathrm{2}\pi{H}\left({a}+{b}\right) \\ $$$$\Rightarrow{f}\left({a},{b}\right)\:=\:{f}_{\mathrm{1}} \left({a}\right)+{f}_{\mathrm{2}} \left({b}\right)−\mathrm{2}\pi\left({a}+{b}\right){H}\left({a}+{b}\right) \\ $$$${f}\left({a},{b}\right)={f}\left({b},{a}\right)\:\:\:\:\left({obvious}\right) \\ $$$$\Rightarrow{f}_{\mathrm{1}} \left({a}\right)={f}_{\mathrm{2}} \left({a}\right)+{c},\:{c}\:{is}\:{just}\:{a}\:{number} \\ $$$$\Rightarrow{f}\left({a},{b}\right)={g}\left({a}\right)+{g}\left({b}\right)−\mathrm{2}\pi\left({a}+{b}\right){H}\left({a}+{b}\right) \\ $$$${f}\left({a},\mathrm{0}\right)=\mathrm{0}={g}\left({a}\right)+{g}\left(\mathrm{0}\right)−\mathrm{2}\pi{aH}\left({a}\right) \\ $$$$\Rightarrow{g}\left({a}\right)=\mathrm{2}\pi{aH}\left({a}\right)−{g}\left(\mathrm{0}\right) \\ $$$$\Rightarrow\mathrm{2}{g}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{g}\left({a}\right)=\mathrm{2}\pi{aH}\left({a}\right) \\ $$$$\Rightarrow\int_{−\infty} ^{\:\infty} …\:=\:\mathrm{2}\pi\left(\mathrm{aH}\left(\mathrm{a}\right)+\mathrm{bH}\left(\mathrm{b}\right)−\left(\mathrm{a}+\mathrm{b}\right)\mathrm{H}\left(\mathrm{a}+\mathrm{b}\right)\right) \\ $$

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