A few ways to integrate $e^{-ax}\sin(x)$:
1) Integration by parts:
$$\begin{align}\int e^{-ax}\sin(x)~dx&=-e^{-ax}\cos(x)-a\int e^{-ax}\cos(x)~dx\\&=-e^{-ax}\cos(x)-a\left(e^{-ax}\sin(x)+a^2\int e^{-ax}\sin(x)~dx\right)\end{align}$$
Let $I=\int e^{-ax}\sin(x)~dx$ to see that
$$I=-e^{-ax}\cos(x)-a\left(e^{-ax}\sin(x)+a^2I\right)$$
which is a linear equation to solve for $I$.
2) Euler's formula:
This is a more complex method (get the pun?) but pretty straight forward. One may either use
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\qquad or\qquad\sin(x)=\Im(e^{ix})$$
Using the second one for simplicity, we see that
$$\begin{align}I&=\Im\int e^{-ax}e^{ix}~dx\\&=\Im\int e^{(i-a)x}~dx\\&=\Im\left(\frac1{i-a}e^{(i-a)x}\right)+c\\&=\Im(u+vi)+c\\&=v+c\end{align}$$
where $v$ is the imaginary part of $\frac1{i-a}e^{(i-a)x}$.
As per the original problem, this is how I would've tackled it, using the complex method:
$$\begin{align}\int_0^1\frac{e^{-ax}\sin(x)}x\ dx&=\int_0^1e^{-ax}\sin(x)\int_0^\infty e^{-xt}\ dt\ dx\\&=\int_0^\infty\int_0^1e^{-(a+t)x}\sin(x)\ dx\ dt\\&=\int_0^\infty\Im\int_0^1e^{[i-(a+t)]x}\ dx\ dt\\&=\int_0^\infty\Im\left(\frac1{i-(a+t)}e^{[i-(a+t)]x}\bigg|_{x=0}^1\right)\ dt\\&=\int_0^\infty\frac1{1+(a+t)^2}\left(1-\frac{\cos(t)+(a+t)\sin(t)}{e^{a+t}}\right)\ dt\end{align}$$
And I think this is far as you can go this way.
Differentiation under integral signs, better known as the Feynman’s trick, is not a standard integration technique taught in curriculum calculus. Nevertheless, it is widely utilized outside classrooms and may appear somewhat magic to those seeing it the first time.
Despite the mystique around it, it is actually rooted in double integrals. A bare-bone illustrative example is
$$I=\int_0^1\int_0^1 x^t dt \>dx= \ln2$$
The natural approach is to integrate $x$ first and then $t$. But, an unsuspecting person may integrate $t$ first and then encounter
$$I=\int_0^1\frac{x-1}{\ln x} dx$$
Now, he/she is stuck since there is not easy way out. Fortunately, there is, which is to differentiate $J(t)$ below under the integral, i.e.
$$J(t)=\int_0^1\frac{x^t-1}{\ln x} dx,\>\>J(t)' = \int_0^1 x^t dx= \frac{1}{1+t}
\implies I=\int_0^1 J(t)'dt=\ln 2$$
A knowledgeable math person, aware of its double-integral origin, would just undo the $t$-integral to reintroduce the double form, and then integrate in the right order,
$$I=\int_0^1\frac{x-1}{\ln x} dx=\int_0^1\int_0^1 x^t dt dx = \int_0^1 \frac1{t+1}dt= \ln 2$$
The two approaches are in fact equivalent, with the double-integrals actually more straightforward. The Feynman’s trick is appealing, since it ‘decouples’ a double-integral in appearance, especially when the embedded double-integral is not immediately discernible. When a seemingly difficult integral is encountered, the differentiation trick is often employed to transform the original integrand to a manageable one.
As a practical example, the trick can be used for deriving the well-known integral
$$I=\int_0^\infty \frac{\sin x}x dx=\frac\pi2$$
with $J(t)=\int_0^\infty \frac{\sin x}x e^{-tx}dx$, $
J’(t)=-\frac{1}{1+t^2}$ and $I=- \int_0^\infty J’(t)dt$.
Best Answer
Differentiate under the integral sign $n$ times as follows $$\int_0^\infty x^n e^{-\lambda x} dx=(-1)^n\frac{d^n}{d\lambda^n} \int_0^\infty e^{-\lambda x} dx= (-1)^n\frac{d^n}{d\lambda^n}\frac1\lambda=\frac{n!}{\lambda^{n+1}} $$