As part of Exploring Quantum Physics - I need to evaluate this integral. Turn out an exercise I did sometimes ago give this as a special case.
An intermediate result of that exercise is this:
$ \begin{eqnarray*} \int{t^ne^{-at}dt} &=& (\frac{1}{a})^n n! \frac{e^{-at}}{-a} -\frac{1}{a}e^{-at}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{a^k(n-k)!} \end{eqnarray*} $
Given this result, we substitute $ a = -i $ , we can simplify this to:
$ \begin{eqnarray*} \int{t^ne^{it}dt} &=& (\frac{1}{-i})^n n! \frac{e^{it}}{i} -\frac{1}{-i}e^{it}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{(-i)^k(n-k)!} \\ &=& i^{n-1} n! e^{it} + e^{it}\sum\limits_{k=0}^{n-1} i^{k+3}\frac{n!t^{n-k}}{(n-k)!} \end{eqnarray*} $
In particular, we are interested in $ n = 1 $ and $ n = 2 $.
$ \begin{eqnarray*} \int{t^1e^{it}dt} &=& i^{1-1} 1! e^{it} + e^{it}\sum\limits_{k=0}^{1-1} i^{k+3}\frac{1!t^{1-k}}{(1-k)!} \\ &=& e^{it} + e^{it} i^{0+3}\frac{1!t^{1-0}}{(1-0)!} \\ &=& e^{it} - ite^{it} \\ &=& e^{it}(-it + 1) \\ \int{t^2e^{it}dt} &=& i^{n-1} n! e^{it} + e^{it}\sum\limits_{k=0}^{n-1} i^{k+3}\frac{n!t^{n-k}}{(n-k)!} \\ &=& i^{2-1} 2! e^{it} + e^{it}\sum\limits_{k=0}^{2-1} i^{k+3}\frac{2!t^{2-k}}{(2-k)!} \\ &=& 2 i e^{it} + e^{it}(i^{0+3}\frac{2!t^{2-0}}{(2-0)!} + i^{1+3}\frac{2!t^{2-1}}{(2-1)!}) \\ &=& 2 i e^{it} + e^{it}(-it^2 + 2t) \\ &=& e^{it}(-it^2 + 2t + 2i) \end{eqnarray*} $
Using Euler's identity we can now write:
$ \begin{eqnarray*} \int{te^{it}dt} &=& e^{it}(-it + it) \\ \int{t(\cos t + i\sin t)dt} &=& (\cos t + i\sin t)(-it + 1) \\ &=& (t\sin t + \cos t) + i(-t\cos t + \sin t) \\ \int{t^2e^{it}dt} &=& e^{it}(-it^2 + 2t + 2i) \\ \int{t^2(\cos t + i\sin t)dt} &=& (\cos t + i\sin t)(-it^2 + 2t + 2i) \\ &=& (t^2 \sin t + 2t \cos t - 2 \sin t) + i(-t^2 \cos t + 2t \sin t + 2 \cos t) \\ \end{eqnarray*} $
Equating the real parts and imaginary parts we get these equations:
An intermediate result of that exercise is this:
$ \begin{eqnarray*} \int{t^ne^{-at}dt} &=& (\frac{1}{a})^n n! \frac{e^{-at}}{-a} -\frac{1}{a}e^{-at}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{a^k(n-k)!} \end{eqnarray*} $
Given this result, we substitute $ a = -i $ , we can simplify this to:
$ \begin{eqnarray*} \int{t^ne^{it}dt} &=& (\frac{1}{-i})^n n! \frac{e^{it}}{i} -\frac{1}{-i}e^{it}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{(-i)^k(n-k)!} \\ &=& i^{n-1} n! e^{it} + e^{it}\sum\limits_{k=0}^{n-1} i^{k+3}\frac{n!t^{n-k}}{(n-k)!} \end{eqnarray*} $
In particular, we are interested in $ n = 1 $ and $ n = 2 $.
$ \begin{eqnarray*} \int{t^1e^{it}dt} &=& i^{1-1} 1! e^{it} + e^{it}\sum\limits_{k=0}^{1-1} i^{k+3}\frac{1!t^{1-k}}{(1-k)!} \\ &=& e^{it} + e^{it} i^{0+3}\frac{1!t^{1-0}}{(1-0)!} \\ &=& e^{it} - ite^{it} \\ &=& e^{it}(-it + 1) \\ \int{t^2e^{it}dt} &=& i^{n-1} n! e^{it} + e^{it}\sum\limits_{k=0}^{n-1} i^{k+3}\frac{n!t^{n-k}}{(n-k)!} \\ &=& i^{2-1} 2! e^{it} + e^{it}\sum\limits_{k=0}^{2-1} i^{k+3}\frac{2!t^{2-k}}{(2-k)!} \\ &=& 2 i e^{it} + e^{it}(i^{0+3}\frac{2!t^{2-0}}{(2-0)!} + i^{1+3}\frac{2!t^{2-1}}{(2-1)!}) \\ &=& 2 i e^{it} + e^{it}(-it^2 + 2t) \\ &=& e^{it}(-it^2 + 2t + 2i) \end{eqnarray*} $
Using Euler's identity we can now write:
$ \begin{eqnarray*} \int{te^{it}dt} &=& e^{it}(-it + it) \\ \int{t(\cos t + i\sin t)dt} &=& (\cos t + i\sin t)(-it + 1) \\ &=& (t\sin t + \cos t) + i(-t\cos t + \sin t) \\ \int{t^2e^{it}dt} &=& e^{it}(-it^2 + 2t + 2i) \\ \int{t^2(\cos t + i\sin t)dt} &=& (\cos t + i\sin t)(-it^2 + 2t + 2i) \\ &=& (t^2 \sin t + 2t \cos t - 2 \sin t) + i(-t^2 \cos t + 2t \sin t + 2 \cos t) \\ \end{eqnarray*} $
Equating the real parts and imaginary parts we get these equations:
- $ \int{t \cos t dt} = t\sin t + \cos t $
- $ \int{t \sin t dt} = -t \cos t + \sin t $
- $ \int{t^2 \cos t dt} = t^2 \sin t + 2t \cos t - 2 \sin t $
- $ \int{t^2 \sin t dt} = -t^2 \cos t + 2t \sin t + 2 \cos t $
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