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EDIT: I think all errors are fixed now. It's 1:30 AM and I'm going to bed now.
Hello,
I am trying to find the Laplace transform of tan(t), but I don't know if I'm getting anywhere. I can't find it in Laplace transform tables and WolframAlpha gives me an answer in terms of complex numbers, hyperbolic trig functions and the digamma function and I don't understand how they get that.
Here is my working and how I got to where I am currently;
[tex]\mathcal{L}\{{\tan}t\} = \int_{0}^{\infty}e^{-st}{\tan}tdt[/tex]
Because by integrating tan(t) I had the feeling I would end up with increasingly complicated integrations if I integrated by parts multiple times so I wrote it as the integral of three functions;
[tex]\int_{0}^{\infty}e^{-st}{\sin}t\frac{1}{{\cos}t}dt[/tex]
Because this is the product of 3 functions I derived what the formula for integration by parts using the product rule like this;
[tex]\frac{d}{dt}uvw = w(udv + vdu) + uvdw[/tex]
Integrating both sides;
[tex]uvw = \int uwdv - \int vwdu - \int uvdw[/tex]
Re-arranging;
[tex]\int uvdw = uvw - \int uwdv - \int vwdu[/tex]
where u, v, w are all functions of t.
For integration by parts I let;
[itex]u = \frac{1}{{\cos}t}[/itex] so [itex]du = \frac{{\sin}t}{{\cos}^{2}t}dt[/itex]
[itex]v = {\sin}t[/itex] so [itex]dv = {\cos}tdt[/itex]
[itex]dw = e^{-st}dt[/itex] so [itex]w = -\frac{1}{s}e^{-st}[/itex]
Using our formula for integration by parts we get;
[tex]-\frac{1}{s}e^{-st}{\sin}t\frac{1}{{\cos}t} - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{1}{{\cos}t}{{\cos}t}dt - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{{\sin}^{2}t}{{\cos}^{2}t}dt[/tex]
Simplifying;
[tex]-\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s}\mathcal{L}\{1\} + \frac{1}{s}\mathcal{L}\{{\tan}^{2}t\}[/tex]
Laplace transform of 1 is just [itex]\frac{1}{s}[/itex], so;
[tex]-\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s^{2}} + \frac{1}{s}\int_{0}^{\infty}e^{-st}{\tan}^{2}tdt[/tex]
Doing integration by parts again, I rewrite the integral as [itex]\int_{0}^{\infty}e^{-st}{\sin}^{2}t\frac{1}{{\cos}^{2}t}dt[/itex].
I let;
[itex]u_{2} = \frac{1}{{\cos}^{2}t}[/itex] so [itex]du_{2} = \frac{{\sin}2t}{{\cos}^{4}t}dt[/itex]
[itex]v_{2} = {\sin}^{2}t[/itex] so [itex]dv_{2} = {\sin}2tdt[/itex]
[itex]dw_{2} = e^{-st}dt[/itex] so [itex]w_{2} = -\frac{1}{s}e^{-st}[/itex]
Using our formula for integration by parts again we get;
[tex]\mathcal{L}\{{\tan}t\} = -\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s^{2}} + \frac{1}{s}\bigg(-\frac{1}{s}e^{-st}\frac{1}{{\cos}^{2}t}{\sin}^{2}t - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{1}{{\cos}^{2}t}{\sin}2tdt - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{{\sin}2t}{{\cos}^{4}t}{\sin}^{2}tdt\bigg)[/tex]
Simplifying;
[tex]\mathcal{L}\{{\tan}t\} = -\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s^{2}} + \frac{1}{s}\bigg(-\frac{1}{s}e^{-st}{\tan}^{2}t + \frac{2}{s}\mathcal{L}\{{\tan}t\} + \frac{2}{s}\mathcal{L}\{{\tan}^{3}t\}\bigg)[/tex]
So as you can see I end up with a tan(t) cubed term to take the Laplace transform of which looks worrying as it looks like no matter how many Laplace transforms I take I'll still end up with some power of tan(t) to take the Laplace transform of, which gets me nowhere. I was wondering if anyone could help me provide a better method of doing this as I think I'll be doing integration by parts forever.
Thanks.
Hello,
I am trying to find the Laplace transform of tan(t), but I don't know if I'm getting anywhere. I can't find it in Laplace transform tables and WolframAlpha gives me an answer in terms of complex numbers, hyperbolic trig functions and the digamma function and I don't understand how they get that.
Here is my working and how I got to where I am currently;
[tex]\mathcal{L}\{{\tan}t\} = \int_{0}^{\infty}e^{-st}{\tan}tdt[/tex]
Because by integrating tan(t) I had the feeling I would end up with increasingly complicated integrations if I integrated by parts multiple times so I wrote it as the integral of three functions;
[tex]\int_{0}^{\infty}e^{-st}{\sin}t\frac{1}{{\cos}t}dt[/tex]
Because this is the product of 3 functions I derived what the formula for integration by parts using the product rule like this;
[tex]\frac{d}{dt}uvw = w(udv + vdu) + uvdw[/tex]
Integrating both sides;
[tex]uvw = \int uwdv - \int vwdu - \int uvdw[/tex]
Re-arranging;
[tex]\int uvdw = uvw - \int uwdv - \int vwdu[/tex]
where u, v, w are all functions of t.
For integration by parts I let;
[itex]u = \frac{1}{{\cos}t}[/itex] so [itex]du = \frac{{\sin}t}{{\cos}^{2}t}dt[/itex]
[itex]v = {\sin}t[/itex] so [itex]dv = {\cos}tdt[/itex]
[itex]dw = e^{-st}dt[/itex] so [itex]w = -\frac{1}{s}e^{-st}[/itex]
Using our formula for integration by parts we get;
[tex]-\frac{1}{s}e^{-st}{\sin}t\frac{1}{{\cos}t} - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{1}{{\cos}t}{{\cos}t}dt - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{{\sin}^{2}t}{{\cos}^{2}t}dt[/tex]
Simplifying;
[tex]-\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s}\mathcal{L}\{1\} + \frac{1}{s}\mathcal{L}\{{\tan}^{2}t\}[/tex]
Laplace transform of 1 is just [itex]\frac{1}{s}[/itex], so;
[tex]-\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s^{2}} + \frac{1}{s}\int_{0}^{\infty}e^{-st}{\tan}^{2}tdt[/tex]
Doing integration by parts again, I rewrite the integral as [itex]\int_{0}^{\infty}e^{-st}{\sin}^{2}t\frac{1}{{\cos}^{2}t}dt[/itex].
I let;
[itex]u_{2} = \frac{1}{{\cos}^{2}t}[/itex] so [itex]du_{2} = \frac{{\sin}2t}{{\cos}^{4}t}dt[/itex]
[itex]v_{2} = {\sin}^{2}t[/itex] so [itex]dv_{2} = {\sin}2tdt[/itex]
[itex]dw_{2} = e^{-st}dt[/itex] so [itex]w_{2} = -\frac{1}{s}e^{-st}[/itex]
Using our formula for integration by parts again we get;
[tex]\mathcal{L}\{{\tan}t\} = -\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s^{2}} + \frac{1}{s}\bigg(-\frac{1}{s}e^{-st}\frac{1}{{\cos}^{2}t}{\sin}^{2}t - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{1}{{\cos}^{2}t}{\sin}2tdt - \int_{0}^{\infty}-\frac{1}{s}e^{-st}\frac{{\sin}2t}{{\cos}^{4}t}{\sin}^{2}tdt\bigg)[/tex]
Simplifying;
[tex]\mathcal{L}\{{\tan}t\} = -\frac{1}{s}e^{-st}{\tan}t + \frac{1}{s^{2}} + \frac{1}{s}\bigg(-\frac{1}{s}e^{-st}{\tan}^{2}t + \frac{2}{s}\mathcal{L}\{{\tan}t\} + \frac{2}{s}\mathcal{L}\{{\tan}^{3}t\}\bigg)[/tex]
So as you can see I end up with a tan(t) cubed term to take the Laplace transform of which looks worrying as it looks like no matter how many Laplace transforms I take I'll still end up with some power of tan(t) to take the Laplace transform of, which gets me nowhere. I was wondering if anyone could help me provide a better method of doing this as I think I'll be doing integration by parts forever.
Thanks.
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