Mixing
Show that the integral does not converge
$begingroup$
Show that $int_0^infty sin(x) ,$ does not converge.
$$int_0^k sin(x) leq int_0^k , dx = k$$
$$lim_{k to infty} k = infty$$
So this integral does not converge.
Is this correct?
calculus
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
$endgroup$
add a comment |
$begingroup$
Show that $int_0^infty sin(x) ,$ does not converge.
$$int_0^k sin(x) leq int_0^k , dx = k$$
$$lim_{k to infty} k = infty$$
So this integral does not converge.
Is this correct?
calculus
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
$endgroup$
$begingroup$
No. It shows that your limit is $leq infty$
$endgroup$
– ThePortakal
May 1 '16 at 13:43
$begingroup$
@ThePortakal So what would the correct solution look like?
$endgroup$
– Si.0788
May 1 '16 at 13:45
$begingroup$
No the integral doesn't converge
$endgroup$
– Archis Welankar
May 1 '16 at 13:46
add a comment |
$begingroup$
Show that $int_0^infty sin(x) ,$ does not converge.
$$int_0^k sin(x) leq int_0^k , dx = k$$
$$lim_{k to infty} k = infty$$
So this integral does not converge.
Is this correct?
calculus
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
$endgroup$
Show that $int_0^infty sin(x) ,$ does not converge.
$$int_0^k sin(x) leq int_0^k , dx = k$$
$$lim_{k to infty} k = infty$$
So this integral does not converge.
Is this correct?
calculus
calculus
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
edited May 1 '16 at 13:41
Si.0788
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
asked May 1 '16 at 13:32
Si.0788Si.0788
1,302527
1,302527
$begingroup$
No. It shows that your limit is $leq infty$
$endgroup$
– ThePortakal
May 1 '16 at 13:43
$begingroup$
@ThePortakal So what would the correct solution look like?
$endgroup$
– Si.0788
May 1 '16 at 13:45
$begingroup$
No the integral doesn't converge
$endgroup$
– Archis Welankar
May 1 '16 at 13:46
add a comment |
$begingroup$
No. It shows that your limit is $leq infty$
$endgroup$
– ThePortakal
May 1 '16 at 13:43
$begingroup$
@ThePortakal So what would the correct solution look like?
$endgroup$
– Si.0788
May 1 '16 at 13:45
$begingroup$
No the integral doesn't converge
$endgroup$
– Archis Welankar
May 1 '16 at 13:46
$begingroup$
No. It shows that your limit is $leq infty$
$endgroup$
– ThePortakal
May 1 '16 at 13:43
$begingroup$
No. It shows that your limit is $leq infty$
$endgroup$
– ThePortakal
May 1 '16 at 13:43
$begingroup$
@ThePortakal So what would the correct solution look like?
$endgroup$
– Si.0788
May 1 '16 at 13:45
$begingroup$
@ThePortakal So what would the correct solution look like?
$endgroup$
– Si.0788
May 1 '16 at 13:45
$begingroup$
No the integral doesn't converge
$endgroup$
– Archis Welankar
May 1 '16 at 13:46
$begingroup$
No the integral doesn't converge
$endgroup$
– Archis Welankar
May 1 '16 at 13:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
big hint
$$int_0^ksin(x)mathrm d x=1-cos(k)$$
Then $int_0^infty sin(x)mathrm d x$ converge if and only if $lim_{kto infty }cos(k)$ exist.
answered May 1 '16 at 13:47
MSEMSE
1,529415
$endgroup$
add a comment |
$begingroup$
It has already been shown in another posted solution that $int_0^infty sin(x),dx$ fails to converge as an improper Riemann Integral.
And clearly we have $int_0^infty |sin(x)|,dx=infty$ and hence $int_0^{infty}sin(x),dx$ fails to exist as a Lebesgue integral.
However, if we interpret the integral as a Fourier Sine Transform then writing the sign function as $text{sgn}$ and the Heaviside Function as $H$ gives
$$begin{align}
int_0^infty sin(x),dx&=frac1{2i}int_{-infty}^{infty}H(x)(e^{ix}-e^{-ix}),dx\\
&=frac1{2i}int_{-infty}^{infty} text{sgn}(x) e^{ix},dx \\
&=frac1{2i} mathscr{F}{text{sgn}}(1)\\
&=1
end{align}$$
So, while $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral, if we more broadly interpret the integral as a DISTRIBUTION, then we can write
$$int_0^infty sin(x),dxsim1$$
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Because you didn't answer the actual question?
$endgroup$
– Math_QED
Jan 12 at 23:52
$begingroup$
@math_qed Read again! I wrote "So, while the integral $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral ..." So what on earth are you talking about?
$endgroup$
– Mark Viola
Jan 13 at 0:50
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:09
$begingroup$
For me the answer is ok! I upvoted it in fact. Happy new year too.
$endgroup$
– Math_QED
Jan 30 at 6:10
$begingroup$
@math_qed Thank you! Much appreciated.
$endgroup$
– Mark Viola
Jan 30 at 15:18
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
big hint
$$int_0^ksin(x)mathrm d x=1-cos(k)$$
Then $int_0^infty sin(x)mathrm d x$ converge if and only if $lim_{kto infty }cos(k)$ exist.
answered May 1 '16 at 13:47
MSEMSE
1,529415
$endgroup$
add a comment |
$begingroup$
big hint
$$int_0^ksin(x)mathrm d x=1-cos(k)$$
Then $int_0^infty sin(x)mathrm d x$ converge if and only if $lim_{kto infty }cos(k)$ exist.
answered May 1 '16 at 13:47
MSEMSE
1,529415
$endgroup$
add a comment |
$begingroup$
big hint
$$int_0^ksin(x)mathrm d x=1-cos(k)$$
Then $int_0^infty sin(x)mathrm d x$ converge if and only if $lim_{kto infty }cos(k)$ exist.
answered May 1 '16 at 13:47
MSEMSE
1,529415
$endgroup$
big hint
$$int_0^ksin(x)mathrm d x=1-cos(k)$$
Then $int_0^infty sin(x)mathrm d x$ converge if and only if $lim_{kto infty }cos(k)$ exist.
answered May 1 '16 at 13:47
MSEMSE
1,529415
answered May 1 '16 at 13:47
MSEMSE
1,529415
answered May 1 '16 at 13:47
MSEMSE
1,529415
answered May 1 '16 at 13:47
MSEMSE
1,529415
1,529415
add a comment |
add a comment |
$begingroup$
It has already been shown in another posted solution that $int_0^infty sin(x),dx$ fails to converge as an improper Riemann Integral.
And clearly we have $int_0^infty |sin(x)|,dx=infty$ and hence $int_0^{infty}sin(x),dx$ fails to exist as a Lebesgue integral.
However, if we interpret the integral as a Fourier Sine Transform then writing the sign function as $text{sgn}$ and the Heaviside Function as $H$ gives
$$begin{align}
int_0^infty sin(x),dx&=frac1{2i}int_{-infty}^{infty}H(x)(e^{ix}-e^{-ix}),dx\\
&=frac1{2i}int_{-infty}^{infty} text{sgn}(x) e^{ix},dx \\
&=frac1{2i} mathscr{F}{text{sgn}}(1)\\
&=1
end{align}$$
So, while $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral, if we more broadly interpret the integral as a DISTRIBUTION, then we can write
$$int_0^infty sin(x),dxsim1$$
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Because you didn't answer the actual question?
$endgroup$
– Math_QED
Jan 12 at 23:52
$begingroup$
@math_qed Read again! I wrote "So, while the integral $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral ..." So what on earth are you talking about?
$endgroup$
– Mark Viola
Jan 13 at 0:50
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:09
$begingroup$
For me the answer is ok! I upvoted it in fact. Happy new year too.
$endgroup$
– Math_QED
Jan 30 at 6:10
$begingroup$
@math_qed Thank you! Much appreciated.
$endgroup$
– Mark Viola
Jan 30 at 15:18
add a comment |
$begingroup$
It has already been shown in another posted solution that $int_0^infty sin(x),dx$ fails to converge as an improper Riemann Integral.
And clearly we have $int_0^infty |sin(x)|,dx=infty$ and hence $int_0^{infty}sin(x),dx$ fails to exist as a Lebesgue integral.
However, if we interpret the integral as a Fourier Sine Transform then writing the sign function as $text{sgn}$ and the Heaviside Function as $H$ gives
$$begin{align}
int_0^infty sin(x),dx&=frac1{2i}int_{-infty}^{infty}H(x)(e^{ix}-e^{-ix}),dx\\
&=frac1{2i}int_{-infty}^{infty} text{sgn}(x) e^{ix},dx \\
&=frac1{2i} mathscr{F}{text{sgn}}(1)\\
&=1
end{align}$$
So, while $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral, if we more broadly interpret the integral as a DISTRIBUTION, then we can write
$$int_0^infty sin(x),dxsim1$$
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Because you didn't answer the actual question?
$endgroup$
– Math_QED
Jan 12 at 23:52
$begingroup$
@math_qed Read again! I wrote "So, while the integral $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral ..." So what on earth are you talking about?
$endgroup$
– Mark Viola
Jan 13 at 0:50
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:09
$begingroup$
For me the answer is ok! I upvoted it in fact. Happy new year too.
$endgroup$
– Math_QED
Jan 30 at 6:10
$begingroup$
@math_qed Thank you! Much appreciated.
$endgroup$
– Mark Viola
Jan 30 at 15:18
add a comment |
$begingroup$
It has already been shown in another posted solution that $int_0^infty sin(x),dx$ fails to converge as an improper Riemann Integral.
And clearly we have $int_0^infty |sin(x)|,dx=infty$ and hence $int_0^{infty}sin(x),dx$ fails to exist as a Lebesgue integral.
However, if we interpret the integral as a Fourier Sine Transform then writing the sign function as $text{sgn}$ and the Heaviside Function as $H$ gives
$$begin{align}
int_0^infty sin(x),dx&=frac1{2i}int_{-infty}^{infty}H(x)(e^{ix}-e^{-ix}),dx\\
&=frac1{2i}int_{-infty}^{infty} text{sgn}(x) e^{ix},dx \\
&=frac1{2i} mathscr{F}{text{sgn}}(1)\\
&=1
end{align}$$
So, while $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral, if we more broadly interpret the integral as a DISTRIBUTION, then we can write
$$int_0^infty sin(x),dxsim1$$
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
$endgroup$
It has already been shown in another posted solution that $int_0^infty sin(x),dx$ fails to converge as an improper Riemann Integral.
And clearly we have $int_0^infty |sin(x)|,dx=infty$ and hence $int_0^{infty}sin(x),dx$ fails to exist as a Lebesgue integral.
However, if we interpret the integral as a Fourier Sine Transform then writing the sign function as $text{sgn}$ and the Heaviside Function as $H$ gives
$$begin{align}
int_0^infty sin(x),dx&=frac1{2i}int_{-infty}^{infty}H(x)(e^{ix}-e^{-ix}),dx\\
&=frac1{2i}int_{-infty}^{infty} text{sgn}(x) e^{ix},dx \\
&=frac1{2i} mathscr{F}{text{sgn}}(1)\\
&=1
end{align}$$
So, while $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral, if we more broadly interpret the integral as a DISTRIBUTION, then we can write
$$int_0^infty sin(x),dxsim1$$
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
edited Jan 13 at 4:50
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
answered Jan 12 at 19:59
Mark ViolaMark Viola
132k1275173
132k1275173
$begingroup$
Because you didn't answer the actual question?
$endgroup$
– Math_QED
Jan 12 at 23:52
$begingroup$
@math_qed Read again! I wrote "So, while the integral $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral ..." So what on earth are you talking about?
$endgroup$
– Mark Viola
Jan 13 at 0:50
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:09
$begingroup$
For me the answer is ok! I upvoted it in fact. Happy new year too.
$endgroup$
– Math_QED
Jan 30 at 6:10
$begingroup$
@math_qed Thank you! Much appreciated.
$endgroup$
– Mark Viola
Jan 30 at 15:18
add a comment |
$begingroup$
Because you didn't answer the actual question?
$endgroup$
– Math_QED
Jan 12 at 23:52
$begingroup$
@math_qed Read again! I wrote "So, while the integral $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral ..." So what on earth are you talking about?
$endgroup$
– Mark Viola
Jan 13 at 0:50
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:09
$begingroup$
For me the answer is ok! I upvoted it in fact. Happy new year too.
$endgroup$
– Math_QED
Jan 30 at 6:10
$begingroup$
@math_qed Thank you! Much appreciated.
$endgroup$
– Mark Viola
Jan 30 at 15:18
$begingroup$
Because you didn't answer the actual question?
$endgroup$
– Math_QED
Jan 12 at 23:52
$begingroup$
Because you didn't answer the actual question?
$endgroup$
– Math_QED
Jan 12 at 23:52
$begingroup$
@math_qed Read again! I wrote "So, while the integral $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral ..." So what on earth are you talking about?
$endgroup$
– Mark Viola
Jan 13 at 0:50
$begingroup$
@math_qed Read again! I wrote "So, while the integral $int_0^infty sin(x),dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral ..." So what on earth are you talking about?
$endgroup$
– Mark Viola
Jan 13 at 0:50
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:09
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
$endgroup$
– Mark Viola
Jan 30 at 5:09
$begingroup$
For me the answer is ok! I upvoted it in fact. Happy new year too.
$endgroup$
– Math_QED
Jan 30 at 6:10
$begingroup$
For me the answer is ok! I upvoted it in fact. Happy new year too.
$endgroup$
– Math_QED
Jan 30 at 6:10
$begingroup$
@math_qed Thank you! Much appreciated.
$endgroup$
– Mark Viola
Jan 30 at 15:18
$begingroup$
@math_qed Thank you! Much appreciated.
$endgroup$
– Mark Viola
Jan 30 at 15:18
add a comment |
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$begingroup$
No. It shows that your limit is $leq infty$
$endgroup$
– ThePortakal
May 1 '16 at 13:43
$begingroup$
@ThePortakal So what would the correct solution look like?
$endgroup$
– Si.0788
May 1 '16 at 13:45
$begingroup$
No the integral doesn't converge
$endgroup$
– Archis Welankar
May 1 '16 at 13:46