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Can someone show me how do I continue from here? Or did I do wrong?

\begin{align} \int \cos(3 \pi t) e^{-jwt} dt &= e^{-jwt} \left(\frac{\sin (3 \pi t)}{3 \pi} \right)-\frac{jw}{(3 \pi)^2}e^{-jwt}\cos(3 \pi t) + \frac{jw}{3 \pi} \int \cos(3 \pi t) e^{-jwt} dt \end{align}

Let $L = ∫ \cos(3πt) e^{-jwt} dt $, then

\begin{align} L &= e^{-jwt} \left(\frac{\sin (3 \pi t)}{3 \pi} \right)-\frac{jw}{(3 \pi)^2}e^{-jwt}\cos(3 \pi t) + \frac{jw}{3 \pi} L \\ \implies L - \frac{jw}{3 \pi} L &= e^{-jwt} \left(\frac{\sin (3 \pi t)}{3 \pi} \right)-\frac{jw}{(3 \pi)^2}e^{-jwt}\cos(3 \pi t) \\ \implies L \left(1 - \frac{jw}{3 \pi} \right) &= \frac{jw}{(3 \pi)^2}\left[\frac{jw}{3 \pi} e^{-jwt}\sin (3 \pi t)- e^{-jwt}\cos(3 \pi t) \right] \\ \implies L &= \frac{1}{\left(1 - \frac{jw}{3 \pi} \right)} \cdot\frac{jw}{(3 \pi)^2} \left[\frac{jw}{3 \pi} e^{-jwt}\sin (3 \pi t)- e^{-jwt}\cos(3 \pi t) \right] \end{align}

Matthew Cassell
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BEX
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  • Check your calculations again. – Nosrati Nov 26 '17 at 10:26
  • You made some algebra mistakes from the second to third lines $$e^{-jwt} \left(\frac{\sin (3 \pi t)}{3 \pi} \right)-\frac{jw}{(3 \pi)^2}e^{-jwt}\cos(3 \pi t) \ne \frac{jw}{(3 \pi)^{2}} \left[ \frac{jw}{3 \pi} e^{-jwt} \sin(3 \pi t) - e^{-jwt} \cos(3 \pi t) \right]$$ – Matthew Cassell Nov 26 '17 at 11:00

1 Answers1

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Well, solving a more general problem:

$$\mathscr{I}_{\space\text{n}}\left(\text{m}\right):=\int\cos\left(\text{n}\cdot x\right)\cdot\exp\left(\text{m}\cdot x\right)\space\text{d}x\tag1$$

Using integration by parts:

$$\int\text{f}\space\text{d}\text{g}=\text{f}\cdot\text{g}-\int\text{g}\space\text{d}\text{f}\tag2$$

Where:

$$\text{f}=\cos\left(\text{n}\cdot x\right)\space,\space\text{d}\text{g}=\exp\left(\text{m}\cdot x\right)\space,\space\text{d}\text{f}=-\text{n}\cdot\sin\left(\text{n}\cdot x\right)\space,\space\text{g}=\frac{\exp\left(\text{m}\cdot x\right)}{\text{m}}\tag3$$

So, we get:

$$\mathscr{I}_{\space\text{n}}\left(\text{m}\right)=\frac{\exp\left(\text{m}\cdot x\right)}{\text{m}}\cdot\cos\left(\text{n}\cdot x\right)+\frac{\text{n}}{\text{m}}\cdot\int\sin\left(\text{n}\cdot x\right)\cdot\exp\left(\text{m}\cdot x\right)\space\text{d}x\tag4$$

Using integration by parts, again:

$$\int\text{q}\space\text{d}\text{p}=\text{q}\cdot\text{p}-\int\text{p}\space\text{d}\text{q}\tag5$$

Where:

$$\text{q}=\sin\left(\text{n}\cdot x\right)\space,\space\text{d}\text{p}=\exp\left(\text{m}\cdot x\right)\space,\space\text{d}\text{q}=\text{n}\cdot\cos\left(\text{n}\cdot x\right)\space,\space\text{p}=\frac{\exp\left(\text{m}\cdot x\right)}{\text{m}}\tag6$$

So, we get:

$$\mathscr{I}_{\space\text{n}}\left(\text{m}\right)=\frac{\text{n}\cdot\sin\left(\text{n}\cdot x\right)\cdot\exp\left(\text{m}\cdot x\right)}{\text{m}^2}+\frac{\cos\left(\text{n}\cdot x\right)\cdot\exp\left(\text{m}\cdot x\right)}{\text{m}}-\frac{\text{n}^2}{\text{m}^2}\cdot\mathscr{I}_{\space\text{n}}\left(\text{m}\right)\tag7$$

So, we also know:

$$\mathscr{I}_{\space\text{n}}\left(\text{m}\right)=\frac{\frac{\text{n}\cdot\sin\left(\text{n}\cdot x\right)\cdot\exp\left(\text{m}\cdot x\right)}{\text{m}^2}+\frac{\cos\left(\text{n}\cdot x\right)\cdot\exp\left(\text{m}\cdot x\right)}{\text{m}}}{1+\frac{\text{n}^2}{\text{m}^2}}\tag8$$

Jan Eerland
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