A twist on an old problem

We will need the following result: $\displaystyle \int x^3 e^{ax} dx = \left(\dfrac{1}{a}x^3-\dfrac{3}{a^2}x^2+\dfrac{6}{a^3}x-\dfrac{6}{a^4}\right)e^{ax}+C$ , and the identity $\cos^2 x = \dfrac{1}{2}+\dfrac{1}{2}\cos(2x)$ .

Let $\displaystyle I = \dfrac{1}{2} \int_0^\pi e^x x^3 dx$ and $\displaystyle J=\dfrac{1}{2} \int_0^\pi e^x x^3 \cos(2x) dx$ , so the answer will be $I+J$ .

We have:

$\displaystyle I = \dfrac{1}{2}\left[(x^3-3x^2+6x-6)e^x \right] \Big|_0^\pi = \dfrac{1}{2}\left[e^\pi(\pi^3-3\pi^2+6\pi-6)+6\right]$

$\displaystyle J = \dfrac{1}{2} \text{Re} \left[ \int_0^\pi e^x x^3 e^{2xi} dx \right] = \dfrac{1}{2} \text{Re} \left[ \int_0^\pi x^3 e^{(1+2i)x} dx \right]$

Let $a=1+2i$ , then, in polar form, $a=\lvert a \rvert e^{\theta i}$ . Then: $\begin{aligned} J &= \dfrac{1}{2} \text{Re} \left[ \left( \dfrac{x^3}{a}-\dfrac{3x^2}{a^2}+\dfrac{6x}{a^3}-\dfrac{6}{a^4} \right) e^{(1+2i)x} \right] \Big|_0^\pi\\ &= \dfrac{1}{2|a|^8} \text{Re} \left[ \left( |a|^7 e^{-\theta i} x^3-3|a|^6 e^{-2\theta i} x^2+6|a|^5 e^{-3\theta i} x-6|a|^4 e^{-4\theta i} \right) e^{(1+2i)x} \right] \Big|_0^\pi\\ &= \dfrac{1}{2|a|^8} \text{Re} \left[ \left( |a|^7 e^{(2x-\theta) i} x^3-3|a|^6 e^{(2x-2\theta) i} x^2+6|a|^5 e^{(2x-3\theta) i} x-6|a|^4 e^{(2x-4\theta) i} \right) e^x \right] \Big|_0^\pi\\ &= \dfrac{1}{2|a|^8} \left[ \left( |a|^7 \cos(2x-\theta) x^3-3|a|^6 \cos(2x-2\theta) x^2+6|a|^5 \cos(2x-3\theta) x-6|a|^4 \cos(2x-4\theta) \right) e^x \right] \Big|_0^\pi\\ &= \dfrac{1}{2|a|^8} \left[ \left( |a|^7 \cos(\theta) \pi^3-3|a|^6 \cos(2\theta) \pi^2+6|a|^5 \cos(3\theta) \pi-6|a|^4 \cos(4\theta) \right) e^\pi + 6|a|^4 \cos(4\theta) \right] \end{aligned}$ But $a^2=-3+4i$ , $a^3=-11-2i$ and $a^4=-7-24i$ , so $\cos(\theta)=\dfrac{1}{|a|}$ , $\cos(2\theta)=-\dfrac{3}{|a|^2}$ , $\cos(3\theta)=-\dfrac{11}{|a|^3}$ and $\cos(4\theta)=-\dfrac{7}{|a|^4}$ . Then: $\begin{aligned} J &= \dfrac{1}{2|a|^8} \left[ \left( |a|^6 \pi^3-3|a|^4 (-3) \pi^2+6|a|^2 (-11) \pi-6(-7) \right) e^\pi + 6(-7) \right]\\ &= \dfrac{1}{1250} \left[ \left(125\pi^3+225\pi^2-330\pi+42 \right) e^\pi - 42 \right] \end{aligned}$ Finally: $I+J = \dfrac{3}{625}\left[ 618+e^\pi(125\pi^3-275\pi^2+570\pi-618) \right]$ . Then, $a=618$ , $b=125$ , $c=-275$ , $d=570$ and $a+b+c+d=\boxed{1038}$ .