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I am going to generalize the result in my post

$$\int xe^x\cos xdx =\frac{e^{x}}{2}[x \cos x+(x-1) \sin x]+C$$ by finding a reduction formula for

$$I_n=\int x^ne^x\cos xdx .$$

After trying for couple of hours, I found that it is hard to evaluate without its partner integral $$J_n=\int x^ne^x\sin xdx$$ Modified Version

As advised by Hans and Martin, I try to find the reduction formula (3) and (4) by complex numbers.

$$ \begin{aligned} I_{n}+i J_{n} &=\int x^{n} e^{x}(\cos x+i \sin x) d x \\ &=\int x^{n} e^{x} \cdot e^{x i} d x \\ &=\int x^{n} e^{x(1+i)} d x \\ &=\frac{1}{1+i} \int x^{n} d\left(e^{(1+i) x}\right) \\ &=\frac{1}{1+i}\left[x^{n} e^{(1+i) x}-\int n x^{n-1} e^{(1+i) x} d x\right] \\ &=\frac{1}{1+i}\left[x^{n} e^{(1+i) x}-n\left(I_{n-1}+i J_{n-1}\right)\right] \\ &=\frac{1-i}{2}\left[x^{n} e^{x}(\cos x+i \sin x)-n\left(I_{n-1}+i J_{n-1}\right)\right] \end{aligned} $$

Now comparing the real and imaginary parts on both sides yields $$ \begin{array}{l} I_{n}=\frac{1}{2}\left[x^{n} e^{x}(\cos x+\sin x)-n\left(I_{n-1}+J_{n-1}\right)\right] \\ J_{n}=\frac{1}{2}\left[x^{n} e^{x}(\sin x-\cos x)+n\left(I_{n-1}-J_{n-1}\right)\right] \end{array} $$

***Original Method ***

First of all, we need to evaluate the following integrals using integration by parts. $$ I_0=\int e^{x} \cos x d x=\frac{e^{x}}{2}(\cos x+\sin x)+c_{1} $$ and $$ J_0=\int e^{x} \sin x d x=\frac{e^{x}}{2}(\sin x-\cos x)+c_{2} $$ Consequently, $$ \int e^{x}(\cos x +\sin x) d x=e^{x} \sin x+c_3 $$ and $$ \int e^{x}(\cos x-\sin x) d x=e^{x} \cos x+c_4 $$ We then obtain easily$$ \begin{aligned} I_{n}+J_{n} &=\int x^{n} \cdot e^{x}(\cos x+\sin x) d x \\ &=\int x^{n} d\left(e^{x} \sin x\right) \\ &=x^{n} e^{x} \sin x-n \int x^{n-1} e^{x} \sin x d x \\ &=x^{n} e^{x} \sin x-n J_{n-1}\qquad \qquad\cdots (1) \end{aligned} $$

Similarly, $$ I_{n}-J_{n}=x^{n} e^{x} \cos x-n I_{n-1} \qquad\qquad \cdots(2) $$ $(1)+(2) $ yields $$ \boxed{I_{n}=\frac{1}{2}\left[x^{n} e^{x}(\cos x+\sin x)-n\left(I_{n-1}+J_{n-1}\right)\right]} \qquad \cdots (3) $$ $(1)-(2) $ yields $$\boxed{J_{n}=\frac{1}{2}\left[x^{n} e^{x}(\sin x-\cos x)+n\left(I_{n-1}-J_{n-1}\right)\right]}\qquad \cdots (4) $$

Now let’s try to find $I_2 $using $ (3) $ and $ (4).$

Using (3) yields $$ \begin{aligned} I_{1} &=\frac{1}{2}\left[x e^{x}(\cos x+\sin x)-\left(I_{0}+J_{0}\right)\right] \\ &=\frac{e^{x}}{2}[x \cos x+(x-1) \sin x]+C_1 \end{aligned} $$ Using (4) yields $$ J_{1}=\frac{e^{x}}{2}\left[x(\sin x-\cos x)+ \cos x\right]+C_2 $$

Using (3) again yields $$ \begin{aligned} I_{2} &=\frac{1}{2}\left[x^{2} e^{x}(\cos x+\sin x)-2\left(I_{1}+J_{1}\right)\right] \\ &=\frac{1}{2}\left[x^{2} e^{x} \cos x+x^{2} e^{x} \sin x-e^{x}(2 x \sin x-\sin x-\cos x)\right]\\&= \frac{e^{x}(x-1)}{2}[(x-1) \sin x+(x+1) \cos x]+C_3 \end{aligned} $$

Using (4) again yields $$ J_{2}=\frac{e^{x}(x-1)}{2}[(x+1) \sin x-(x-1) \cos x]+C_{4} $$

My Question

Is there other alternative methods? You opinions and solutions are highly appreciated.

Lai
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    I haven't checked the details, but perhaps you could first compute $\int x^n e^{zx} dx$ and then let $z=1+i$ and take real and imaginary parts? – Hans Lundmark Mar 03 '22 at 08:42
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    And for $\int x^n e^{zx} dx$, see for example https://math.stackexchange.com/q/21516/42969 and https://math.stackexchange.com/q/1501073/42969. – Martin R Mar 03 '22 at 08:52
  • For what it’s worth, there is an explicit formula for $I_s$ being $I_s = -\frac{1}{2} x^{s+1} \left(E_{-s}( (-i-1)x)+E_{-s} ((i-1)x)\right)+C$, where $E_{-s}$ is the generalised exponential integral function. For $s \in \mathbb{N}$, the expression reduces into simple sinusoidal functions as you found. – KStar Mar 03 '22 at 08:55
  • Thank you for your advices which are absorbed in the modified version. – Lai Mar 03 '22 at 14:55

1 Answers1

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Note that

\begin{align} \int x^n e^{ax}dx =\frac{d^n}{da^n}\int e^{ax}dx =\frac{d^n}{da^n}\left(\frac{e^{ax}}a\right) = \sum_{k=0}^n \frac{(-1)^k n!}{(n-k)!} \frac{x^{n-k}e^{ax}}{a^{k+1}} \end{align} Then \begin{align} \int x^n e^{x}\cos x\>dx =&\>\Re \int x^n e^{(1+i)x}dx =\>\Re\sum_{k=0}^n \frac{(-1)^k n!}{(n-k)!} \frac{x^{n-k}e^{(1+i)x}}{(1+i)^{k+1}}\\ =&\sum_{k=0}^n \frac{(-1)^k n! x^{n-k}e^{x}}{2^{\frac{k+1}2}(n-k)!} \cos\bigg(x - \frac{(k+1)\pi}4 \bigg) \\ \\ \int x^n e^{x}\sin x\>dx =&\sum_{k=0}^n \frac{(-1)^k n! x^{n-k}e^{x}}{2^{\frac{k+1}2}(n-k)!} \sin\bigg(x - \frac{(k+1)\pi}4 \bigg) \end{align}

Quanto
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