Mixing
integrating $sin(px)sin(qx)/x^2$
Show that the integral from $0$ to infinity of $sin(px)sin(qx)/x^2$ equals $picdotmin(p,q)/2$, where $p,q>0$.
I need to use Cauchy's Residue theorem i think, but I can't see what function to apply it to since the integrand in the question has only removable singularities. Also, I'm struggling to see where the $min(p,q)$ term in the solution comes from.
Can anyone help me solve this question? Thanks
complex-analysis
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
add a comment |
Show that the integral from $0$ to infinity of $sin(px)sin(qx)/x^2$ equals $picdotmin(p,q)/2$, where $p,q>0$.
I need to use Cauchy's Residue theorem i think, but I can't see what function to apply it to since the integrand in the question has only removable singularities. Also, I'm struggling to see where the $min(p,q)$ term in the solution comes from.
Can anyone help me solve this question? Thanks
complex-analysis
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
add a comment |
Show that the integral from $0$ to infinity of $sin(px)sin(qx)/x^2$ equals $picdotmin(p,q)/2$, where $p,q>0$.
I need to use Cauchy's Residue theorem i think, but I can't see what function to apply it to since the integrand in the question has only removable singularities. Also, I'm struggling to see where the $min(p,q)$ term in the solution comes from.
Can anyone help me solve this question? Thanks
complex-analysis
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
Show that the integral from $0$ to infinity of $sin(px)sin(qx)/x^2$ equals $picdotmin(p,q)/2$, where $p,q>0$.
I need to use Cauchy's Residue theorem i think, but I can't see what function to apply it to since the integrand in the question has only removable singularities. Also, I'm struggling to see where the $min(p,q)$ term in the solution comes from.
Can anyone help me solve this question? Thanks
complex-analysis
complex-analysis
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
edited Dec 2 '13 at 12:45
Harald Hanche-Olsen
27.7k24064
27.7k24064
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
asked Dec 2 '13 at 12:43
DocMartinDocMartin
112
112
add a comment |
add a comment |
1 Answer
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The integrand is even, so integrate over the whole real line and divide by $2$.
You could try replacing one of the sine terms (say, $sin(px)$) with $-ie^{ipx}$. The real part is still what you want, and the imaginary part gives rise to an integral with an odd integrand, so that should give a zero. Except there is now a pole at the origin. Play around with semicircles, one large and one small centered at the origin, and joined by paths along the real axis.
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
add a comment |
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1 Answer
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1 Answer
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The integrand is even, so integrate over the whole real line and divide by $2$.
You could try replacing one of the sine terms (say, $sin(px)$) with $-ie^{ipx}$. The real part is still what you want, and the imaginary part gives rise to an integral with an odd integrand, so that should give a zero. Except there is now a pole at the origin. Play around with semicircles, one large and one small centered at the origin, and joined by paths along the real axis.
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
add a comment |
The integrand is even, so integrate over the whole real line and divide by $2$.
You could try replacing one of the sine terms (say, $sin(px)$) with $-ie^{ipx}$. The real part is still what you want, and the imaginary part gives rise to an integral with an odd integrand, so that should give a zero. Except there is now a pole at the origin. Play around with semicircles, one large and one small centered at the origin, and joined by paths along the real axis.
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
add a comment |
The integrand is even, so integrate over the whole real line and divide by $2$.
You could try replacing one of the sine terms (say, $sin(px)$) with $-ie^{ipx}$. The real part is still what you want, and the imaginary part gives rise to an integral with an odd integrand, so that should give a zero. Except there is now a pole at the origin. Play around with semicircles, one large and one small centered at the origin, and joined by paths along the real axis.
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
The integrand is even, so integrate over the whole real line and divide by $2$.
You could try replacing one of the sine terms (say, $sin(px)$) with $-ie^{ipx}$. The real part is still what you want, and the imaginary part gives rise to an integral with an odd integrand, so that should give a zero. Except there is now a pole at the origin. Play around with semicircles, one large and one small centered at the origin, and joined by paths along the real axis.
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
answered Dec 2 '13 at 12:50
Harald Hanche-OlsenHarald Hanche-Olsen
27.7k24064
27.7k24064
add a comment |
add a comment |
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