Mixing
Calculate the integral using complex analysis $int_0^{infty}frac{sin(sqrt s)}{(1+s)^{2}} ds$
$begingroup$
The task is to calculate this integral using complex analysis: I know some methods, which involve calculating the integral around half of the circle, also it would be wise to add here $i cos(s^{frac{1}{2}})$ and then take the real part of the result. But here comes my problem, this function isn't even, so I am unable to move the limit 0 to $-infty$, which makes it problematic, because integrating over one fourth of the circle involves integrating along the imaginary line, which is not simpler than this. Any help appreciated.
$$int_0^{infty}frac{sin(s^{frac{1}{2}})}{(1+s)^{2}} ds$$
complex-analysis complex-integration
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
$endgroup$
add a comment |
$begingroup$
The task is to calculate this integral using complex analysis: I know some methods, which involve calculating the integral around half of the circle, also it would be wise to add here $i cos(s^{frac{1}{2}})$ and then take the real part of the result. But here comes my problem, this function isn't even, so I am unable to move the limit 0 to $-infty$, which makes it problematic, because integrating over one fourth of the circle involves integrating along the imaginary line, which is not simpler than this. Any help appreciated.
$$int_0^{infty}frac{sin(s^{frac{1}{2}})}{(1+s)^{2}} ds$$
complex-analysis complex-integration
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
$endgroup$
1
$begingroup$
You may want to first substitute $s=u^2$.
$endgroup$
– Zachary
Feb 1 at 18:23
$begingroup$
Wow right thank you!
$endgroup$
– ryszard eggink
Feb 1 at 18:25
add a comment |
$begingroup$
The task is to calculate this integral using complex analysis: I know some methods, which involve calculating the integral around half of the circle, also it would be wise to add here $i cos(s^{frac{1}{2}})$ and then take the real part of the result. But here comes my problem, this function isn't even, so I am unable to move the limit 0 to $-infty$, which makes it problematic, because integrating over one fourth of the circle involves integrating along the imaginary line, which is not simpler than this. Any help appreciated.
$$int_0^{infty}frac{sin(s^{frac{1}{2}})}{(1+s)^{2}} ds$$
complex-analysis complex-integration
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
$endgroup$
The task is to calculate this integral using complex analysis: I know some methods, which involve calculating the integral around half of the circle, also it would be wise to add here $i cos(s^{frac{1}{2}})$ and then take the real part of the result. But here comes my problem, this function isn't even, so I am unable to move the limit 0 to $-infty$, which makes it problematic, because integrating over one fourth of the circle involves integrating along the imaginary line, which is not simpler than this. Any help appreciated.
$$int_0^{infty}frac{sin(s^{frac{1}{2}})}{(1+s)^{2}} ds$$
complex-analysis complex-integration
complex-analysis complex-integration
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
edited Feb 1 at 19:32
user625055
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
asked Feb 1 at 18:17
ryszard egginkryszard eggink
420110
420110
1
$begingroup$
You may want to first substitute $s=u^2$.
$endgroup$
– Zachary
Feb 1 at 18:23
$begingroup$
Wow right thank you!
$endgroup$
– ryszard eggink
Feb 1 at 18:25
add a comment |
1
$begingroup$
You may want to first substitute $s=u^2$.
$endgroup$
– Zachary
Feb 1 at 18:23
$begingroup$
Wow right thank you!
$endgroup$
– ryszard eggink
Feb 1 at 18:25
1
1
$begingroup$
You may want to first substitute $s=u^2$.
$endgroup$
– Zachary
Feb 1 at 18:23
$begingroup$
You may want to first substitute $s=u^2$.
$endgroup$
– Zachary
Feb 1 at 18:23
$begingroup$
Wow right thank you!
$endgroup$
– ryszard eggink
Feb 1 at 18:25
$begingroup$
Wow right thank you!
$endgroup$
– ryszard eggink
Feb 1 at 18:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Integrating by parts gives
$$int_0^{infty}frac{sin(sqrt s)}{(1+s)^{2}} ds=int_0^infty frac{cos (sqrt s)}{1+s}frac{ds}{2sqrt s}$$
Now substitution $s=x^2$ and the integral become
$$int_0^infty frac{cos x}{1+x^2}dx$$
Since the integrand is an even function and the real part of $e^{ix}$ is $cos x$ we have the following equality
$$int_0^infty frac{cos x}{1+x^2}dx=frac12int_{-infty}^infty frac{e^{ix}}{(x+i)(x-i)}dx$$
Choosing the contour to be the upper semi-plane where only the pole $z=i$ exists for the function $$f(z)=frac{e^{iz}}{(z+i)(z-i)}$$
Give the integral to be $$frac12 cdot 2pi i lim_{zto i} cdot (z-i)cdot text{Res} f(z)=pi i lim_{zto i} frac{e^{iz}}{z+i} =pi i frac{e^{-1}}{2i}=frac{pi}{2e}$$
$endgroup$
1
$begingroup$
Thanks, i got the same results after a hint from the comment :)
$endgroup$
– ryszard eggink
Feb 1 at 19:30
add a comment |
$begingroup$
$$newcommand{Res}{operatorname*{Res}}
begin{align}
int_0^inftyfrac{sinleft(s^{1/2}right)}{(1+s)^2},mathrm{d}s
&=int_0^inftyfrac{2ssin(s)}{(1+s^2)^2},mathrm{d}stag1\
&=int_{-infty}^inftyfrac{ssin(s)}{(1+s^2)^2},mathrm{d}stag2\
&=int_{-infty}^inftyfrac1{2i}left(e^{is}-e^{-is}right)frac1{4i}left(frac1{(s-i)^2}-frac1{(s+i)^2}right)mathrm{d}stag3\
&=-frac18left[2pi iRes_{s=i}left(frac{e^{is}}{(s-i)^2}right)-2pi iRes_{s=-i}left(frac{e^{-is}}{(s+i)^2}right)right]tag4\[6pt]
&=-frac18left[-frac{2pi}e-frac{2pi}eright]tag5\[9pt]
&=fracpi{2e}tag6
end{align}
$$
Explanation:
$(1)$: substitute $smapsto s^2$
$(2)$: use symmetry
$(3)$: write $sin(x)$ as exponentials and apply partial fractions
$(4)$: use the contour $[-R,R]cup Re^{i[0,pi]}$ for $e^{is}$ which contains the singularity at $s=i$
$phantom{(4)text{:}}$ use the contour $[-R,R]cup Re^{i[0,-pi]}$ for $e^{-is}$ which contains the singularity at $s=-i$
$phantom{(4)text{:}}$ note that the second contour is clockwise
$(5)$: evaluate the residues
$(6)$: simplify
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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$begingroup$
Integrating by parts gives
$$int_0^{infty}frac{sin(sqrt s)}{(1+s)^{2}} ds=int_0^infty frac{cos (sqrt s)}{1+s}frac{ds}{2sqrt s}$$
Now substitution $s=x^2$ and the integral become
$$int_0^infty frac{cos x}{1+x^2}dx$$
Since the integrand is an even function and the real part of $e^{ix}$ is $cos x$ we have the following equality
$$int_0^infty frac{cos x}{1+x^2}dx=frac12int_{-infty}^infty frac{e^{ix}}{(x+i)(x-i)}dx$$
Choosing the contour to be the upper semi-plane where only the pole $z=i$ exists for the function $$f(z)=frac{e^{iz}}{(z+i)(z-i)}$$
Give the integral to be $$frac12 cdot 2pi i lim_{zto i} cdot (z-i)cdot text{Res} f(z)=pi i lim_{zto i} frac{e^{iz}}{z+i} =pi i frac{e^{-1}}{2i}=frac{pi}{2e}$$
$endgroup$
1
$begingroup$
Thanks, i got the same results after a hint from the comment :)
$endgroup$
– ryszard eggink
Feb 1 at 19:30
add a comment |
$begingroup$
Integrating by parts gives
$$int_0^{infty}frac{sin(sqrt s)}{(1+s)^{2}} ds=int_0^infty frac{cos (sqrt s)}{1+s}frac{ds}{2sqrt s}$$
Now substitution $s=x^2$ and the integral become
$$int_0^infty frac{cos x}{1+x^2}dx$$
Since the integrand is an even function and the real part of $e^{ix}$ is $cos x$ we have the following equality
$$int_0^infty frac{cos x}{1+x^2}dx=frac12int_{-infty}^infty frac{e^{ix}}{(x+i)(x-i)}dx$$
Choosing the contour to be the upper semi-plane where only the pole $z=i$ exists for the function $$f(z)=frac{e^{iz}}{(z+i)(z-i)}$$
Give the integral to be $$frac12 cdot 2pi i lim_{zto i} cdot (z-i)cdot text{Res} f(z)=pi i lim_{zto i} frac{e^{iz}}{z+i} =pi i frac{e^{-1}}{2i}=frac{pi}{2e}$$
$endgroup$
1
$begingroup$
Thanks, i got the same results after a hint from the comment :)
$endgroup$
– ryszard eggink
Feb 1 at 19:30
add a comment |
$begingroup$
Integrating by parts gives
$$int_0^{infty}frac{sin(sqrt s)}{(1+s)^{2}} ds=int_0^infty frac{cos (sqrt s)}{1+s}frac{ds}{2sqrt s}$$
Now substitution $s=x^2$ and the integral become
$$int_0^infty frac{cos x}{1+x^2}dx$$
Since the integrand is an even function and the real part of $e^{ix}$ is $cos x$ we have the following equality
$$int_0^infty frac{cos x}{1+x^2}dx=frac12int_{-infty}^infty frac{e^{ix}}{(x+i)(x-i)}dx$$
Choosing the contour to be the upper semi-plane where only the pole $z=i$ exists for the function $$f(z)=frac{e^{iz}}{(z+i)(z-i)}$$
Give the integral to be $$frac12 cdot 2pi i lim_{zto i} cdot (z-i)cdot text{Res} f(z)=pi i lim_{zto i} frac{e^{iz}}{z+i} =pi i frac{e^{-1}}{2i}=frac{pi}{2e}$$
$endgroup$
Integrating by parts gives
$$int_0^{infty}frac{sin(sqrt s)}{(1+s)^{2}} ds=int_0^infty frac{cos (sqrt s)}{1+s}frac{ds}{2sqrt s}$$
Now substitution $s=x^2$ and the integral become
$$int_0^infty frac{cos x}{1+x^2}dx$$
Since the integrand is an even function and the real part of $e^{ix}$ is $cos x$ we have the following equality
$$int_0^infty frac{cos x}{1+x^2}dx=frac12int_{-infty}^infty frac{e^{ix}}{(x+i)(x-i)}dx$$
Choosing the contour to be the upper semi-plane where only the pole $z=i$ exists for the function $$f(z)=frac{e^{iz}}{(z+i)(z-i)}$$
Give the integral to be $$frac12 cdot 2pi i lim_{zto i} cdot (z-i)cdot text{Res} f(z)=pi i lim_{zto i} frac{e^{iz}}{z+i} =pi i frac{e^{-1}}{2i}=frac{pi}{2e}$$
edited Feb 1 at 19:34
answered Feb 1 at 19:29
user625055
1
$begingroup$
Thanks, i got the same results after a hint from the comment :)
$endgroup$
– ryszard eggink
Feb 1 at 19:30
add a comment |
1
$begingroup$
Thanks, i got the same results after a hint from the comment :)
$endgroup$
– ryszard eggink
Feb 1 at 19:30
1
1
$begingroup$
Thanks, i got the same results after a hint from the comment :)
$endgroup$
– ryszard eggink
Feb 1 at 19:30
$begingroup$
Thanks, i got the same results after a hint from the comment :)
$endgroup$
– ryszard eggink
Feb 1 at 19:30
add a comment |
$begingroup$
$$newcommand{Res}{operatorname*{Res}}
begin{align}
int_0^inftyfrac{sinleft(s^{1/2}right)}{(1+s)^2},mathrm{d}s
&=int_0^inftyfrac{2ssin(s)}{(1+s^2)^2},mathrm{d}stag1\
&=int_{-infty}^inftyfrac{ssin(s)}{(1+s^2)^2},mathrm{d}stag2\
&=int_{-infty}^inftyfrac1{2i}left(e^{is}-e^{-is}right)frac1{4i}left(frac1{(s-i)^2}-frac1{(s+i)^2}right)mathrm{d}stag3\
&=-frac18left[2pi iRes_{s=i}left(frac{e^{is}}{(s-i)^2}right)-2pi iRes_{s=-i}left(frac{e^{-is}}{(s+i)^2}right)right]tag4\[6pt]
&=-frac18left[-frac{2pi}e-frac{2pi}eright]tag5\[9pt]
&=fracpi{2e}tag6
end{align}
$$
Explanation:
$(1)$: substitute $smapsto s^2$
$(2)$: use symmetry
$(3)$: write $sin(x)$ as exponentials and apply partial fractions
$(4)$: use the contour $[-R,R]cup Re^{i[0,pi]}$ for $e^{is}$ which contains the singularity at $s=i$
$phantom{(4)text{:}}$ use the contour $[-R,R]cup Re^{i[0,-pi]}$ for $e^{-is}$ which contains the singularity at $s=-i$
$phantom{(4)text{:}}$ note that the second contour is clockwise
$(5)$: evaluate the residues
$(6)$: simplify
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
$endgroup$
add a comment |
$begingroup$
$$newcommand{Res}{operatorname*{Res}}
begin{align}
int_0^inftyfrac{sinleft(s^{1/2}right)}{(1+s)^2},mathrm{d}s
&=int_0^inftyfrac{2ssin(s)}{(1+s^2)^2},mathrm{d}stag1\
&=int_{-infty}^inftyfrac{ssin(s)}{(1+s^2)^2},mathrm{d}stag2\
&=int_{-infty}^inftyfrac1{2i}left(e^{is}-e^{-is}right)frac1{4i}left(frac1{(s-i)^2}-frac1{(s+i)^2}right)mathrm{d}stag3\
&=-frac18left[2pi iRes_{s=i}left(frac{e^{is}}{(s-i)^2}right)-2pi iRes_{s=-i}left(frac{e^{-is}}{(s+i)^2}right)right]tag4\[6pt]
&=-frac18left[-frac{2pi}e-frac{2pi}eright]tag5\[9pt]
&=fracpi{2e}tag6
end{align}
$$
Explanation:
$(1)$: substitute $smapsto s^2$
$(2)$: use symmetry
$(3)$: write $sin(x)$ as exponentials and apply partial fractions
$(4)$: use the contour $[-R,R]cup Re^{i[0,pi]}$ for $e^{is}$ which contains the singularity at $s=i$
$phantom{(4)text{:}}$ use the contour $[-R,R]cup Re^{i[0,-pi]}$ for $e^{-is}$ which contains the singularity at $s=-i$
$phantom{(4)text{:}}$ note that the second contour is clockwise
$(5)$: evaluate the residues
$(6)$: simplify
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
$endgroup$
add a comment |
$begingroup$
$$newcommand{Res}{operatorname*{Res}}
begin{align}
int_0^inftyfrac{sinleft(s^{1/2}right)}{(1+s)^2},mathrm{d}s
&=int_0^inftyfrac{2ssin(s)}{(1+s^2)^2},mathrm{d}stag1\
&=int_{-infty}^inftyfrac{ssin(s)}{(1+s^2)^2},mathrm{d}stag2\
&=int_{-infty}^inftyfrac1{2i}left(e^{is}-e^{-is}right)frac1{4i}left(frac1{(s-i)^2}-frac1{(s+i)^2}right)mathrm{d}stag3\
&=-frac18left[2pi iRes_{s=i}left(frac{e^{is}}{(s-i)^2}right)-2pi iRes_{s=-i}left(frac{e^{-is}}{(s+i)^2}right)right]tag4\[6pt]
&=-frac18left[-frac{2pi}e-frac{2pi}eright]tag5\[9pt]
&=fracpi{2e}tag6
end{align}
$$
Explanation:
$(1)$: substitute $smapsto s^2$
$(2)$: use symmetry
$(3)$: write $sin(x)$ as exponentials and apply partial fractions
$(4)$: use the contour $[-R,R]cup Re^{i[0,pi]}$ for $e^{is}$ which contains the singularity at $s=i$
$phantom{(4)text{:}}$ use the contour $[-R,R]cup Re^{i[0,-pi]}$ for $e^{-is}$ which contains the singularity at $s=-i$
$phantom{(4)text{:}}$ note that the second contour is clockwise
$(5)$: evaluate the residues
$(6)$: simplify
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
$endgroup$
$$newcommand{Res}{operatorname*{Res}}
begin{align}
int_0^inftyfrac{sinleft(s^{1/2}right)}{(1+s)^2},mathrm{d}s
&=int_0^inftyfrac{2ssin(s)}{(1+s^2)^2},mathrm{d}stag1\
&=int_{-infty}^inftyfrac{ssin(s)}{(1+s^2)^2},mathrm{d}stag2\
&=int_{-infty}^inftyfrac1{2i}left(e^{is}-e^{-is}right)frac1{4i}left(frac1{(s-i)^2}-frac1{(s+i)^2}right)mathrm{d}stag3\
&=-frac18left[2pi iRes_{s=i}left(frac{e^{is}}{(s-i)^2}right)-2pi iRes_{s=-i}left(frac{e^{-is}}{(s+i)^2}right)right]tag4\[6pt]
&=-frac18left[-frac{2pi}e-frac{2pi}eright]tag5\[9pt]
&=fracpi{2e}tag6
end{align}
$$
Explanation:
$(1)$: substitute $smapsto s^2$
$(2)$: use symmetry
$(3)$: write $sin(x)$ as exponentials and apply partial fractions
$(4)$: use the contour $[-R,R]cup Re^{i[0,pi]}$ for $e^{is}$ which contains the singularity at $s=i$
$phantom{(4)text{:}}$ use the contour $[-R,R]cup Re^{i[0,-pi]}$ for $e^{-is}$ which contains the singularity at $s=-i$
$phantom{(4)text{:}}$ note that the second contour is clockwise
$(5)$: evaluate the residues
$(6)$: simplify
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
edited Feb 2 at 0:21
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
answered Feb 1 at 19:33
robjohn♦robjohn
271k27313642
271k27313642
add a comment |
add a comment |
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1
$begingroup$
You may want to first substitute $s=u^2$.
$endgroup$
– Zachary
Feb 1 at 18:23
$begingroup$
Wow right thank you!
$endgroup$
– ryszard eggink
Feb 1 at 18:25