Mixing
Evaluate the integral $int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx$
$begingroup$
I would like to evaluate the following integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx$$
I have made several attempts using some crude methods, the workout is way too long to format and type out, so I will make it as brief as possible.
Firstly, I evaluated the integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}dx$$
Using integration by parts and the properties of definite integrals:
Setting $u=sin(ax)sin(bx)$ and $dv=x^{-2}$
Using:
- $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$
- $int_{-infty}^infty f(x)dx=frac 1 2int_{0}^infty f(x)dx$
- $2sin acos b=sin(a+b)+sin(a-b)$
I obtained:
$$frac1 2bpi$$
I tried the same approach with the original question, and obtained 3 sets of integrals:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For (1) and (2), the term $a$ cancels out as expected.
I have no idea how (1) and (2) resulted in $arctan(k^{-1})$, I just put the intrgral in wolfram alpha. I think inverse Fourier Transforms were involved here. I also don't think $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$ can be used here, since we have an $e^{-kx}$ term. I went with it just to see where I can get to.
I am unable to solve integral (3).
My second attempt involved dominant and monotone convergence theorems, which was a mess. I don't even know what I was trying to do, hence I will not type it out here. I'm 100% sure it is totally wrong.
We were given some hints:
There's a common and smart trigonometric identity involved, then invoke convergence theorems. The final answer looks very neat.
I suspect the "smart trigonometric identity" is the one listed above, and my answer for the integral without the exponent did indeed look very neat. However using the same method, I wasn't able to obtain the result I needed. It should be a lot easier if convergence theorems are used, at least I assume it would.
I appreciate any help or hints. I'd like to know how this was obtained:
$int_{0}^infty operatorname{sinc} xe^{-kx}dx = arctan(k^{-1})$
Also please pick out any mistakes I've made, as I am certain my method for the question was highly fallacious and/or mathematically unrigorous. I think I'm on the right track though.
Thanks in advance!
measure-theory convergence improper-integrals riemann-integration
edited May 16 '18 at 12:42
Harry Peter
5,49911439
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
$endgroup$
add a comment |
$begingroup$
I would like to evaluate the following integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx$$
I have made several attempts using some crude methods, the workout is way too long to format and type out, so I will make it as brief as possible.
Firstly, I evaluated the integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}dx$$
Using integration by parts and the properties of definite integrals:
Setting $u=sin(ax)sin(bx)$ and $dv=x^{-2}$
Using:
- $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$
- $int_{-infty}^infty f(x)dx=frac 1 2int_{0}^infty f(x)dx$
- $2sin acos b=sin(a+b)+sin(a-b)$
I obtained:
$$frac1 2bpi$$
I tried the same approach with the original question, and obtained 3 sets of integrals:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For (1) and (2), the term $a$ cancels out as expected.
I have no idea how (1) and (2) resulted in $arctan(k^{-1})$, I just put the intrgral in wolfram alpha. I think inverse Fourier Transforms were involved here. I also don't think $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$ can be used here, since we have an $e^{-kx}$ term. I went with it just to see where I can get to.
I am unable to solve integral (3).
My second attempt involved dominant and monotone convergence theorems, which was a mess. I don't even know what I was trying to do, hence I will not type it out here. I'm 100% sure it is totally wrong.
We were given some hints:
There's a common and smart trigonometric identity involved, then invoke convergence theorems. The final answer looks very neat.
I suspect the "smart trigonometric identity" is the one listed above, and my answer for the integral without the exponent did indeed look very neat. However using the same method, I wasn't able to obtain the result I needed. It should be a lot easier if convergence theorems are used, at least I assume it would.
I appreciate any help or hints. I'd like to know how this was obtained:
$int_{0}^infty operatorname{sinc} xe^{-kx}dx = arctan(k^{-1})$
Also please pick out any mistakes I've made, as I am certain my method for the question was highly fallacious and/or mathematically unrigorous. I think I'm on the right track though.
Thanks in advance!
measure-theory convergence improper-integrals riemann-integration
edited May 16 '18 at 12:42
Harry Peter
5,49911439
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
$endgroup$
$begingroup$
Do you know Laplace transform ?
$endgroup$
– C. Dubussy
May 14 '18 at 10:51
$begingroup$
Yes, I do know Laplace transforms.
$endgroup$
– Hypergeometry
May 14 '18 at 10:55
$begingroup$
For the last one, the antiderivative involves the exponential integral function.
$endgroup$
– Claude Leibovici
May 14 '18 at 10:56
$begingroup$
I've never studied exponential integral functions, and I think according to the hint, it isn't required. The answer should look as neat as $frac{1}{2}bpi$. I'll look into that function and see what I can spin out of it though.
$endgroup$
– Hypergeometry
May 14 '18 at 11:00
add a comment |
$begingroup$
I would like to evaluate the following integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx$$
I have made several attempts using some crude methods, the workout is way too long to format and type out, so I will make it as brief as possible.
Firstly, I evaluated the integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}dx$$
Using integration by parts and the properties of definite integrals:
Setting $u=sin(ax)sin(bx)$ and $dv=x^{-2}$
Using:
- $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$
- $int_{-infty}^infty f(x)dx=frac 1 2int_{0}^infty f(x)dx$
- $2sin acos b=sin(a+b)+sin(a-b)$
I obtained:
$$frac1 2bpi$$
I tried the same approach with the original question, and obtained 3 sets of integrals:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For (1) and (2), the term $a$ cancels out as expected.
I have no idea how (1) and (2) resulted in $arctan(k^{-1})$, I just put the intrgral in wolfram alpha. I think inverse Fourier Transforms were involved here. I also don't think $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$ can be used here, since we have an $e^{-kx}$ term. I went with it just to see where I can get to.
I am unable to solve integral (3).
My second attempt involved dominant and monotone convergence theorems, which was a mess. I don't even know what I was trying to do, hence I will not type it out here. I'm 100% sure it is totally wrong.
We were given some hints:
There's a common and smart trigonometric identity involved, then invoke convergence theorems. The final answer looks very neat.
I suspect the "smart trigonometric identity" is the one listed above, and my answer for the integral without the exponent did indeed look very neat. However using the same method, I wasn't able to obtain the result I needed. It should be a lot easier if convergence theorems are used, at least I assume it would.
I appreciate any help or hints. I'd like to know how this was obtained:
$int_{0}^infty operatorname{sinc} xe^{-kx}dx = arctan(k^{-1})$
Also please pick out any mistakes I've made, as I am certain my method for the question was highly fallacious and/or mathematically unrigorous. I think I'm on the right track though.
Thanks in advance!
measure-theory convergence improper-integrals riemann-integration
edited May 16 '18 at 12:42
Harry Peter
5,49911439
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
$endgroup$
I would like to evaluate the following integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx$$
I have made several attempts using some crude methods, the workout is way too long to format and type out, so I will make it as brief as possible.
Firstly, I evaluated the integral:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}dx$$
Using integration by parts and the properties of definite integrals:
Setting $u=sin(ax)sin(bx)$ and $dv=x^{-2}$
Using:
- $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$
- $int_{-infty}^infty f(x)dx=frac 1 2int_{0}^infty f(x)dx$
- $2sin acos b=sin(a+b)+sin(a-b)$
I obtained:
$$frac1 2bpi$$
I tried the same approach with the original question, and obtained 3 sets of integrals:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For (1) and (2), the term $a$ cancels out as expected.
I have no idea how (1) and (2) resulted in $arctan(k^{-1})$, I just put the intrgral in wolfram alpha. I think inverse Fourier Transforms were involved here. I also don't think $int_{-infty}^infty frac{g(ax)}{x}dx = int_{-infty}^infty frac{g(x)}{x}dx$ can be used here, since we have an $e^{-kx}$ term. I went with it just to see where I can get to.
I am unable to solve integral (3).
My second attempt involved dominant and monotone convergence theorems, which was a mess. I don't even know what I was trying to do, hence I will not type it out here. I'm 100% sure it is totally wrong.
We were given some hints:
There's a common and smart trigonometric identity involved, then invoke convergence theorems. The final answer looks very neat.
I suspect the "smart trigonometric identity" is the one listed above, and my answer for the integral without the exponent did indeed look very neat. However using the same method, I wasn't able to obtain the result I needed. It should be a lot easier if convergence theorems are used, at least I assume it would.
I appreciate any help or hints. I'd like to know how this was obtained:
$int_{0}^infty operatorname{sinc} xe^{-kx}dx = arctan(k^{-1})$
Also please pick out any mistakes I've made, as I am certain my method for the question was highly fallacious and/or mathematically unrigorous. I think I'm on the right track though.
Thanks in advance!
measure-theory convergence improper-integrals riemann-integration
measure-theory convergence improper-integrals riemann-integration
edited May 16 '18 at 12:42
Harry Peter
5,49911439
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
edited May 16 '18 at 12:42
Harry Peter
5,49911439
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
edited May 16 '18 at 12:42
Harry Peter
5,49911439
edited May 16 '18 at 12:42
Harry Peter
5,49911439
edited May 16 '18 at 12:42
Harry Peter
5,49911439
5,49911439
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
asked May 14 '18 at 10:46
HypergeometryHypergeometry
1059
1059
$begingroup$
Do you know Laplace transform ?
$endgroup$
– C. Dubussy
May 14 '18 at 10:51
$begingroup$
Yes, I do know Laplace transforms.
$endgroup$
– Hypergeometry
May 14 '18 at 10:55
$begingroup$
For the last one, the antiderivative involves the exponential integral function.
$endgroup$
– Claude Leibovici
May 14 '18 at 10:56
$begingroup$
I've never studied exponential integral functions, and I think according to the hint, it isn't required. The answer should look as neat as $frac{1}{2}bpi$. I'll look into that function and see what I can spin out of it though.
$endgroup$
– Hypergeometry
May 14 '18 at 11:00
add a comment |
$begingroup$
Do you know Laplace transform ?
$endgroup$
– C. Dubussy
May 14 '18 at 10:51
$begingroup$
Yes, I do know Laplace transforms.
$endgroup$
– Hypergeometry
May 14 '18 at 10:55
$begingroup$
For the last one, the antiderivative involves the exponential integral function.
$endgroup$
– Claude Leibovici
May 14 '18 at 10:56
$begingroup$
I've never studied exponential integral functions, and I think according to the hint, it isn't required. The answer should look as neat as $frac{1}{2}bpi$. I'll look into that function and see what I can spin out of it though.
$endgroup$
– Hypergeometry
May 14 '18 at 11:00
$begingroup$
Do you know Laplace transform ?
$endgroup$
– C. Dubussy
May 14 '18 at 10:51
$begingroup$
Do you know Laplace transform ?
$endgroup$
– C. Dubussy
May 14 '18 at 10:51
$begingroup$
Yes, I do know Laplace transforms.
$endgroup$
– Hypergeometry
May 14 '18 at 10:55
$begingroup$
Yes, I do know Laplace transforms.
$endgroup$
– Hypergeometry
May 14 '18 at 10:55
$begingroup$
For the last one, the antiderivative involves the exponential integral function.
$endgroup$
– Claude Leibovici
May 14 '18 at 10:56
$begingroup$
For the last one, the antiderivative involves the exponential integral function.
$endgroup$
– Claude Leibovici
May 14 '18 at 10:56
$begingroup$
I've never studied exponential integral functions, and I think according to the hint, it isn't required. The answer should look as neat as $frac{1}{2}bpi$. I'll look into that function and see what I can spin out of it though.
$endgroup$
– Hypergeometry
May 14 '18 at 11:00
$begingroup$
I've never studied exponential integral functions, and I think according to the hint, it isn't required. The answer should look as neat as $frac{1}{2}bpi$. I'll look into that function and see what I can spin out of it though.
$endgroup$
– Hypergeometry
May 14 '18 at 11:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A common strategy is to write $x^{-n}=frac{1}{Gamma (n)}int_0^infty y^{n-1}e^{-xy}dy$. For example, $$int_0^inftyfrac{sin x}{x}e^{-kx}dx=int_0^infty dyBigg[int_0^infty e^{-(k+y)x}sin x dxBigg]=int_0^inftyfrac{dy}{(k+y)^2+1}.$$
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely.
$endgroup$
– Hypergeometry
May 14 '18 at 13:10
$begingroup$
I computed the integral and obtained $frac{pi b}{2}$. Still don't know how to solve the third integral in the top post question.
$endgroup$
– Hypergeometry
May 14 '18 at 14:28
$begingroup$
@Hypergeometry The integral representation of $1/x$ should work.
$endgroup$
– J.G.
May 14 '18 at 15:31
$begingroup$
I used dominant convergence theorem and solved that part. Easy but annoyingly long.
$endgroup$
– Hypergeometry
May 15 '18 at 4:41
add a comment |
$begingroup$
I've solved the integral, the answer looks odd, and it is very very long. I'll post the results later. It's 5am the I have a class that begins at 10.
Basically, these are all wrong:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For 1 and 2, $(a+b)$ cannot be removed from the integral. The result has the inverse tangent function.
For 3, I used dominant convergence theorem which allowed me to differentiate within the integral. (I checked all conditions and they were satisfied) The result was a set of functions involving $ln(x)$.
Here's the the result looks like:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx=frac{1}{2(a+b)}(frac{pi}{2}-arctan(frac{k}{a+b})+frac{1}{2(b-a)}(arctan(frac{k}{b-a})-frac{pi}{2})+frac{barctan(frac{a-b}{k})}{k}+frac{1}{2}ln((a+b)^2+k^2)-frac{1}{2}ln((a-b)^2+k^2)$$
Which is just ridiculous. I assume it can be simplified into something nicer, or I just made a massive mistake somewhere in my 13 pages of writing. I'll know the result later today anyway. I'll post a proper solution if mine is wrong.
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
$endgroup$
$begingroup$
Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed.
$endgroup$
– Hypergeometry
May 15 '18 at 4:17
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A common strategy is to write $x^{-n}=frac{1}{Gamma (n)}int_0^infty y^{n-1}e^{-xy}dy$. For example, $$int_0^inftyfrac{sin x}{x}e^{-kx}dx=int_0^infty dyBigg[int_0^infty e^{-(k+y)x}sin x dxBigg]=int_0^inftyfrac{dy}{(k+y)^2+1}.$$
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely.
$endgroup$
– Hypergeometry
May 14 '18 at 13:10
$begingroup$
I computed the integral and obtained $frac{pi b}{2}$. Still don't know how to solve the third integral in the top post question.
$endgroup$
– Hypergeometry
May 14 '18 at 14:28
$begingroup$
@Hypergeometry The integral representation of $1/x$ should work.
$endgroup$
– J.G.
May 14 '18 at 15:31
$begingroup$
I used dominant convergence theorem and solved that part. Easy but annoyingly long.
$endgroup$
– Hypergeometry
May 15 '18 at 4:41
add a comment |
$begingroup$
A common strategy is to write $x^{-n}=frac{1}{Gamma (n)}int_0^infty y^{n-1}e^{-xy}dy$. For example, $$int_0^inftyfrac{sin x}{x}e^{-kx}dx=int_0^infty dyBigg[int_0^infty e^{-(k+y)x}sin x dxBigg]=int_0^inftyfrac{dy}{(k+y)^2+1}.$$
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely.
$endgroup$
– Hypergeometry
May 14 '18 at 13:10
$begingroup$
I computed the integral and obtained $frac{pi b}{2}$. Still don't know how to solve the third integral in the top post question.
$endgroup$
– Hypergeometry
May 14 '18 at 14:28
$begingroup$
@Hypergeometry The integral representation of $1/x$ should work.
$endgroup$
– J.G.
May 14 '18 at 15:31
$begingroup$
I used dominant convergence theorem and solved that part. Easy but annoyingly long.
$endgroup$
– Hypergeometry
May 15 '18 at 4:41
add a comment |
$begingroup$
A common strategy is to write $x^{-n}=frac{1}{Gamma (n)}int_0^infty y^{n-1}e^{-xy}dy$. For example, $$int_0^inftyfrac{sin x}{x}e^{-kx}dx=int_0^infty dyBigg[int_0^infty e^{-(k+y)x}sin x dxBigg]=int_0^inftyfrac{dy}{(k+y)^2+1}.$$
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
$endgroup$
A common strategy is to write $x^{-n}=frac{1}{Gamma (n)}int_0^infty y^{n-1}e^{-xy}dy$. For example, $$int_0^inftyfrac{sin x}{x}e^{-kx}dx=int_0^infty dyBigg[int_0^infty e^{-(k+y)x}sin x dxBigg]=int_0^inftyfrac{dy}{(k+y)^2+1}.$$
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
edited Jan 29 at 16:32
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
answered May 14 '18 at 12:08
J.G.J.G.
32.5k23250
32.5k23250
$begingroup$
Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely.
$endgroup$
– Hypergeometry
May 14 '18 at 13:10
$begingroup$
I computed the integral and obtained $frac{pi b}{2}$. Still don't know how to solve the third integral in the top post question.
$endgroup$
– Hypergeometry
May 14 '18 at 14:28
$begingroup$
@Hypergeometry The integral representation of $1/x$ should work.
$endgroup$
– J.G.
May 14 '18 at 15:31
$begingroup$
I used dominant convergence theorem and solved that part. Easy but annoyingly long.
$endgroup$
– Hypergeometry
May 15 '18 at 4:41
add a comment |
$begingroup$
Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely.
$endgroup$
– Hypergeometry
May 14 '18 at 13:10
$begingroup$
I computed the integral and obtained $frac{pi b}{2}$. Still don't know how to solve the third integral in the top post question.
$endgroup$
– Hypergeometry
May 14 '18 at 14:28
$begingroup$
@Hypergeometry The integral representation of $1/x$ should work.
$endgroup$
– J.G.
May 14 '18 at 15:31
$begingroup$
I used dominant convergence theorem and solved that part. Easy but annoyingly long.
$endgroup$
– Hypergeometry
May 15 '18 at 4:41
$begingroup$
Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely.
$endgroup$
– Hypergeometry
May 14 '18 at 13:10
$begingroup$
Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely.
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– Hypergeometry
May 14 '18 at 13:10
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I computed the integral and obtained $frac{pi b}{2}$. Still don't know how to solve the third integral in the top post question.
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– Hypergeometry
May 14 '18 at 14:28
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I computed the integral and obtained $frac{pi b}{2}$. Still don't know how to solve the third integral in the top post question.
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– Hypergeometry
May 14 '18 at 14:28
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@Hypergeometry The integral representation of $1/x$ should work.
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– J.G.
May 14 '18 at 15:31
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@Hypergeometry The integral representation of $1/x$ should work.
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– J.G.
May 14 '18 at 15:31
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I used dominant convergence theorem and solved that part. Easy but annoyingly long.
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– Hypergeometry
May 15 '18 at 4:41
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I used dominant convergence theorem and solved that part. Easy but annoyingly long.
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– Hypergeometry
May 15 '18 at 4:41
add a comment |
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I've solved the integral, the answer looks odd, and it is very very long. I'll post the results later. It's 5am the I have a class that begins at 10.
Basically, these are all wrong:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For 1 and 2, $(a+b)$ cannot be removed from the integral. The result has the inverse tangent function.
For 3, I used dominant convergence theorem which allowed me to differentiate within the integral. (I checked all conditions and they were satisfied) The result was a set of functions involving $ln(x)$.
Here's the the result looks like:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx=frac{1}{2(a+b)}(frac{pi}{2}-arctan(frac{k}{a+b})+frac{1}{2(b-a)}(arctan(frac{k}{b-a})-frac{pi}{2})+frac{barctan(frac{a-b}{k})}{k}+frac{1}{2}ln((a+b)^2+k^2)-frac{1}{2}ln((a-b)^2+k^2)$$
Which is just ridiculous. I assume it can be simplified into something nicer, or I just made a massive mistake somewhere in my 13 pages of writing. I'll know the result later today anyway. I'll post a proper solution if mine is wrong.
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
$endgroup$
$begingroup$
Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed.
$endgroup$
– Hypergeometry
May 15 '18 at 4:17
add a comment |
$begingroup$
I've solved the integral, the answer looks odd, and it is very very long. I'll post the results later. It's 5am the I have a class that begins at 10.
Basically, these are all wrong:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For 1 and 2, $(a+b)$ cannot be removed from the integral. The result has the inverse tangent function.
For 3, I used dominant convergence theorem which allowed me to differentiate within the integral. (I checked all conditions and they were satisfied) The result was a set of functions involving $ln(x)$.
Here's the the result looks like:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx=frac{1}{2(a+b)}(frac{pi}{2}-arctan(frac{k}{a+b})+frac{1}{2(b-a)}(arctan(frac{k}{b-a})-frac{pi}{2})+frac{barctan(frac{a-b}{k})}{k}+frac{1}{2}ln((a+b)^2+k^2)-frac{1}{2}ln((a-b)^2+k^2)$$
Which is just ridiculous. I assume it can be simplified into something nicer, or I just made a massive mistake somewhere in my 13 pages of writing. I'll know the result later today anyway. I'll post a proper solution if mine is wrong.
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
$endgroup$
$begingroup$
Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed.
$endgroup$
– Hypergeometry
May 15 '18 at 4:17
add a comment |
$begingroup$
I've solved the integral, the answer looks odd, and it is very very long. I'll post the results later. It's 5am the I have a class that begins at 10.
Basically, these are all wrong:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For 1 and 2, $(a+b)$ cannot be removed from the integral. The result has the inverse tangent function.
For 3, I used dominant convergence theorem which allowed me to differentiate within the integral. (I checked all conditions and they were satisfied) The result was a set of functions involving $ln(x)$.
Here's the the result looks like:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx=frac{1}{2(a+b)}(frac{pi}{2}-arctan(frac{k}{a+b})+frac{1}{2(b-a)}(arctan(frac{k}{b-a})-frac{pi}{2})+frac{barctan(frac{a-b}{k})}{k}+frac{1}{2}ln((a+b)^2+k^2)-frac{1}{2}ln((a-b)^2+k^2)$$
Which is just ridiculous. I assume it can be simplified into something nicer, or I just made a massive mistake somewhere in my 13 pages of writing. I'll know the result later today anyway. I'll post a proper solution if mine is wrong.
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
$endgroup$
I've solved the integral, the answer looks odd, and it is very very long. I'll post the results later. It's 5am the I have a class that begins at 10.
Basically, these are all wrong:
$frac{1}{2}(a+b)int_{0}^infty operatorname{sinc}xe^{-kx}dx = frac{1}{2}(a+b)arctan(k^{-1})$
$frac{1}{2}(b-a)int_{0}^infty operatorname{sinc} xe^{-kx}dx = frac{1}{2}(b-a)arctan(k^{-1})$
$(-frac{1}{2}k)int_{0}^infty frac {sin(ax)sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)
For 1 and 2, $(a+b)$ cannot be removed from the integral. The result has the inverse tangent function.
For 3, I used dominant convergence theorem which allowed me to differentiate within the integral. (I checked all conditions and they were satisfied) The result was a set of functions involving $ln(x)$.
Here's the the result looks like:
$$int_0^infty frac{sin(ax)}{x} frac{sin(bx)}{x}e^{-kx}dx=frac{1}{2(a+b)}(frac{pi}{2}-arctan(frac{k}{a+b})+frac{1}{2(b-a)}(arctan(frac{k}{b-a})-frac{pi}{2})+frac{barctan(frac{a-b}{k})}{k}+frac{1}{2}ln((a+b)^2+k^2)-frac{1}{2}ln((a-b)^2+k^2)$$
Which is just ridiculous. I assume it can be simplified into something nicer, or I just made a massive mistake somewhere in my 13 pages of writing. I'll know the result later today anyway. I'll post a proper solution if mine is wrong.
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
answered May 14 '18 at 18:58
HypergeometryHypergeometry
1059
1059
$begingroup$
Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed.
$endgroup$
– Hypergeometry
May 15 '18 at 4:17
add a comment |
$begingroup$
Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed.
$endgroup$
– Hypergeometry
May 15 '18 at 4:17
$begingroup$
Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed.
$endgroup$
– Hypergeometry
May 15 '18 at 4:17
$begingroup$
Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed.
$endgroup$
– Hypergeometry
May 15 '18 at 4:17
add a comment |
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Do you know Laplace transform ?
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– C. Dubussy
May 14 '18 at 10:51
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Yes, I do know Laplace transforms.
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– Hypergeometry
May 14 '18 at 10:55
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For the last one, the antiderivative involves the exponential integral function.
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– Claude Leibovici
May 14 '18 at 10:56
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I've never studied exponential integral functions, and I think according to the hint, it isn't required. The answer should look as neat as $frac{1}{2}bpi$. I'll look into that function and see what I can spin out of it though.
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– Hypergeometry
May 14 '18 at 11:00