Question about integral of an odd functionHow can I find the integral of this function using trig...
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Question about integral of an odd function
How can I find the integral of this function using trig substitution?Why does $sin{alpha}cdot isin{alpha x}$ disappear from this integral?Using Polar Coordinates to Calculate Double IntegralQuestion concerning the domain of polar coordinate.Evaluate $int_0^{infty}frac{e^{-x}-e^{-2x}}{x}dx$ using a double integralQuestion about the limits of definite integralsQuestion about a substitution in an integralDoubt about an improper multiple integralIntegrate $int_0^1 sin^{-1}{frac{x^2}{1+x^2}}dx$Studying the convergence of the integral $int_0^pi frac{ln(sin(x))}{x}dx$
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I am studying something and encountered this:
"
Let $R(theta,T) = int_{-T}^{T} frac{(sin theta t)}{t}dt, S(T) = int_0^Tfrac{(sin x)}{x}dx$, then for $theta > 0$ and changing variables $t=x/theta $ shows that
$R(theta,T)=2int_0^{Ttheta}frac{sin x}{x}dx = 2S(Ttheta)$ while for $theta<0$, $R(theta,T) = -R(|theta|,T)$" which I don't understand.
If $theta<0$, then $R(theta,T)=2int_{-T|theta|}^{0}frac{sin x}{x}dx =2int_0^{T|theta|}frac{sin x}{x}dx = R(|theta|,T)$ as $frac{sin x}{x}$ is an even function, right? I am missing something simple here, thanks and appreciate an explanation.
calculus
asked 4 hours ago
manifoldedmanifolded
3487
$endgroup$
add a comment |
$begingroup$
I am studying something and encountered this:
"
Let $R(theta,T) = int_{-T}^{T} frac{(sin theta t)}{t}dt, S(T) = int_0^Tfrac{(sin x)}{x}dx$, then for $theta > 0$ and changing variables $t=x/theta $ shows that
$R(theta,T)=2int_0^{Ttheta}frac{sin x}{x}dx = 2S(Ttheta)$ while for $theta<0$, $R(theta,T) = -R(|theta|,T)$" which I don't understand.
If $theta<0$, then $R(theta,T)=2int_{-T|theta|}^{0}frac{sin x}{x}dx =2int_0^{T|theta|}frac{sin x}{x}dx = R(|theta|,T)$ as $frac{sin x}{x}$ is an even function, right? I am missing something simple here, thanks and appreciate an explanation.
calculus
asked 4 hours ago
manifoldedmanifolded
3487
$endgroup$
add a comment |
$begingroup$
I am studying something and encountered this:
"
Let $R(theta,T) = int_{-T}^{T} frac{(sin theta t)}{t}dt, S(T) = int_0^Tfrac{(sin x)}{x}dx$, then for $theta > 0$ and changing variables $t=x/theta $ shows that
$R(theta,T)=2int_0^{Ttheta}frac{sin x}{x}dx = 2S(Ttheta)$ while for $theta<0$, $R(theta,T) = -R(|theta|,T)$" which I don't understand.
If $theta<0$, then $R(theta,T)=2int_{-T|theta|}^{0}frac{sin x}{x}dx =2int_0^{T|theta|}frac{sin x}{x}dx = R(|theta|,T)$ as $frac{sin x}{x}$ is an even function, right? I am missing something simple here, thanks and appreciate an explanation.
calculus
asked 4 hours ago
manifoldedmanifolded
3487
$endgroup$
I am studying something and encountered this:
"
Let $R(theta,T) = int_{-T}^{T} frac{(sin theta t)}{t}dt, S(T) = int_0^Tfrac{(sin x)}{x}dx$, then for $theta > 0$ and changing variables $t=x/theta $ shows that
$R(theta,T)=2int_0^{Ttheta}frac{sin x}{x}dx = 2S(Ttheta)$ while for $theta<0$, $R(theta,T) = -R(|theta|,T)$" which I don't understand.
If $theta<0$, then $R(theta,T)=2int_{-T|theta|}^{0}frac{sin x}{x}dx =2int_0^{T|theta|}frac{sin x}{x}dx = R(|theta|,T)$ as $frac{sin x}{x}$ is an even function, right? I am missing something simple here, thanks and appreciate an explanation.
calculus
calculus
asked 4 hours ago
manifoldedmanifolded
3487
asked 4 hours ago
manifoldedmanifolded
3487
asked 4 hours ago
manifoldedmanifolded
3487
asked 4 hours ago
manifoldedmanifolded
3487
asked 4 hours ago
manifoldedmanifolded
3487
3487
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When $t=-T$ we get $x=ttheta =-Ttheta =T|theta|$ and not $-T|theta|$.
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
1
$begingroup$
Correct! Thanks.
$endgroup$
– manifolded
4 hours ago
add a comment |
$begingroup$
$$begin{align}theta <0 & Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (theta t)}{t}dt\
&Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (-|theta| t)}{t}dt\
&Rightarrow R(theta ,T)=-int_{-T}^Tfrac{sin (|theta| t)}{t}dt [because sin(-x)=-sin (x)big]\
&Rightarrow R(theta ,T)=-R(|theta|,T)\
end{align}$$
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
When $t=-T$ we get $x=ttheta =-Ttheta =T|theta|$ and not $-T|theta|$.
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
1
$begingroup$
Correct! Thanks.
$endgroup$
– manifolded
4 hours ago
add a comment |
$begingroup$
When $t=-T$ we get $x=ttheta =-Ttheta =T|theta|$ and not $-T|theta|$.
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
1
$begingroup$
Correct! Thanks.
$endgroup$
– manifolded
4 hours ago
add a comment |
$begingroup$
When $t=-T$ we get $x=ttheta =-Ttheta =T|theta|$ and not $-T|theta|$.
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
When $t=-T$ we get $x=ttheta =-Ttheta =T|theta|$ and not $-T|theta|$.
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
answered 4 hours ago
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
63.1k42362
1
$begingroup$
Correct! Thanks.
$endgroup$
– manifolded
4 hours ago
add a comment |
1
$begingroup$
Correct! Thanks.
$endgroup$
– manifolded
4 hours ago
1
1
$begingroup$
Correct! Thanks.
$endgroup$
– manifolded
4 hours ago
$begingroup$
Correct! Thanks.
$endgroup$
– manifolded
4 hours ago
add a comment |
$begingroup$
$$begin{align}theta <0 & Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (theta t)}{t}dt\
&Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (-|theta| t)}{t}dt\
&Rightarrow R(theta ,T)=-int_{-T}^Tfrac{sin (|theta| t)}{t}dt [because sin(-x)=-sin (x)big]\
&Rightarrow R(theta ,T)=-R(|theta|,T)\
end{align}$$
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
$endgroup$
add a comment |
$begingroup$
$$begin{align}theta <0 & Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (theta t)}{t}dt\
&Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (-|theta| t)}{t}dt\
&Rightarrow R(theta ,T)=-int_{-T}^Tfrac{sin (|theta| t)}{t}dt [because sin(-x)=-sin (x)big]\
&Rightarrow R(theta ,T)=-R(|theta|,T)\
end{align}$$
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
$endgroup$
add a comment |
$begingroup$
$$begin{align}theta <0 & Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (theta t)}{t}dt\
&Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (-|theta| t)}{t}dt\
&Rightarrow R(theta ,T)=-int_{-T}^Tfrac{sin (|theta| t)}{t}dt [because sin(-x)=-sin (x)big]\
&Rightarrow R(theta ,T)=-R(|theta|,T)\
end{align}$$
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
$endgroup$
$$begin{align}theta <0 & Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (theta t)}{t}dt\
&Rightarrow R(theta ,T)=int_{-T}^Tfrac{sin (-|theta| t)}{t}dt\
&Rightarrow R(theta ,T)=-int_{-T}^Tfrac{sin (|theta| t)}{t}dt [because sin(-x)=-sin (x)big]\
&Rightarrow R(theta ,T)=-R(|theta|,T)\
end{align}$$
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
answered 4 hours ago
s0ulr3aper07s0ulr3aper07
541111
541111
add a comment |
add a comment |
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