Some properties to choose from if you are having difficulty....
Max Woesner
1. Find ![{\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24cc81a9b57842b359b9c425520ad66afbf2e696)
Recall 
, so ![{\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]={\mathcal {F}}[cos(2\pi f_{0}t)g(t)]=\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77a621df08dc561c1881f885d34a6db5224fa008)
Also recall 
,so ![{\displaystyle \int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }{\frac {1}{2}}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa67da17dc1d9f04fa8b72c6d0cf3a712f40c70f)
Now ![{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{2}}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt={\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f-f_{0})t}g(t)dt+{\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f+f_{0})t}g(t)dt={\frac {1}{2}}G(f-f_{0})+{\frac {1}{2}}G(f+f_{0})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/398e4f99de8f675135c1cdd4dd2564861a179119)
So ![{\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]={\frac {1}{2}}[G(f-f_{0})+G(f+f_{0})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00db7d7e53690eaefe05d548d51d6c52d015e43d)
reviewed by Joshua Sarris
2. Find ![{\displaystyle {\mathcal {F}}{\bigg [}\int _{-\infty }^{\infty }g(t)h^{*}(t)dt{\bigg ]}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ece2c7fc000ccdabd9d2bb3159ba2052fe8c7eb7)
Recall ![{\displaystyle g(t)={\mathcal {F}}^{-1}[G(f)]=\int _{-\infty }^{\infty }G(f)e^{j2\pi ft}df\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ff9253bd7e0c3e6c8d6f016ea2887c4c5407043)
Similarly, ![{\displaystyle h(t)={\mathcal {F}}^{-1}[H(f)]=\int _{-\infty }^{\infty }H(f)e^{j2\pi ft}df\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55a1ea4c246e23b54fcb8f5563be01e266cc31b4)
So ![{\displaystyle {\mathcal {F}}{\bigg [}\int _{-\infty }^{\infty }g(t)h^{*}(t)dt{\bigg ]}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }G(f^{'})e^{j2\pi f^{'}t}df^{'}{\Bigg (}\int _{-\infty }^{\infty }H(f^{''})e^{j2\pi f^{''}t}df^{''}{\Bigg )}^{*}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47d22b81919bf6e034d7b62f802bb1d3344f884f)
Now 
Note that 
So ![{\displaystyle {\mathcal {F}}{\bigg [}\int _{-\infty }^{\infty }g(t)h^{*}(t)dt{\bigg ]}=\int _{-\infty }^{\infty }G(f)H^{*}(f)df\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e172f905a8f394f6a04851d6d4dc457e99a22cce)
Reviewed by Nick Christman
Nick Christman
1. Find ![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c2c3e80b66f61426213cda99d534f8179a11d3a)
To begin, we know that
![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right)e^{-j2\pi ft}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/129708b4ac17bb5999b7cf7e4807b7f001e1566e)
After some factoring and combinting of like terms we get:
![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }10^{t}g(t)\,dt\right)e^{j2\pi f(t_{0}-t)}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ac179d818fd8beba9b44a994cd5bc6d7c713175)
But recall that
Because of this definition (and some "math magic") our problem has been simplified significantly:
![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=\int _{-\infty }^{\infty }10^{t}g(t)\delta (t-t_{0})\,dt=10^{t_{0}}g(t_{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/303a1c0a8e83c2975bf78dd7ace2f2a137d2a22b)
Therefore,
![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=10^{t_{0}}g(t_{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4204cb42d09cf4c46dd6fc2ea9aed8b512ac830)
2.
Joshua Sarris
Find ![{\displaystyle {\mathcal {F}}[sin(w_{0}t)g(t)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/982c9e181460db5faafc47902bc3dd11e92fb7f8)
Recall
,
so expanding we have,
![{\displaystyle {\mathcal {F}}[sin(w_{0}t)g(t)]={\mathcal {F}}[sin(2\pi f_{0}t)g(t)]=\int _{-\infty }^{\infty }sin(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6225ec01a20e83a776ded489642cdb29d069d33c)
Also recall
,
so we can convert to exponentials.
![{\displaystyle \int _{-\infty }^{\infty }sin(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }{\frac {1}{j2}}[e^{j2\pi f_{0}t}-e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f88b7b79b8a8cc5dcc6a94c71ad76dd9feda0422)
Now integrating gives us,
![{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{j2}}[e^{j2\pi f_{0}t}-e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt={\frac {1}{j2}}\int _{-\infty }^{\infty }e^{-j2\pi (f-f_{0})t}g(t)dt-{\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f+f_{0})t}g(t)dt={\frac {1}{j2}}G(f-f_{0})-{\frac {1}{j2}}G(f+f_{0})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56c7af34ce898a8b6ac13e839d33f108437f6433)
So we now have the identity,
![{\displaystyle {\mathcal {F}}[sin(w_{0}t)g(t)]={\frac {1}{j2}}[G(f-f_{0})-G(f+f_{0})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6b41814c468a20025b44e90e113ba5897671c85)
or rather
![{\displaystyle {\mathcal {F}}[sin(w_{0}t)g(t)]={\frac {1}{2}}j[G(f+f_{0})+G(f-f_{0})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a3158a8c8e7e6aafc5a409d50027eb3c330e174)
Reviewed by Max