I've been trying to find solve this integral for a while, using differentiation under the integral sign:
$$\int_0^1 \! \frac{{e^{-ax}}\sin(x)}{x} \, \mathrm{d}x$$
But I keep getting stuck around here, when I'm trying to find the indefinite integral with respect to $x$:
$$-a\int\!e^{-ax}\sin(x) \, \mathrm{d}x$$
How would I go about solving this and am I even on the right track with the second integral?
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.
Following up on the previous answer, any method you chose you should arrive at the same result for the derivative with respect to a.
For notation purposes, let's say $F(a)=\int_0^1 \! \frac{{e^{-ax}}\sin(x)}{x} dx$. Now:
$$F'(a)=-\int_{0}^{1}\!e^{-ax}\sin(x) \, \mathrm{d}x=\frac{e^{-a}(\cos(1)+a\sin(1))}{a^2+1}-\frac{1}{a^2+1}$$
In order to recover $F(x)$ we need to integrate that... The second term integrates to $\tan^{-1}(a)$, but the first term is not solvable in terms of elementary functions, wolfram says it is the following: $$F(a)=\int \frac{e^{-a}(\cos(1)+a\sin(1))}{a^2+1} da=\frac{1}{2}[i \space Ei(-a-i)-i\space Ei(i-a)]+C$$
For the constant term, set $a=0$, and equate to find $C=-\frac{\pi}{2}$, giving$$F(a)=\frac{1}{2}[i \space Ei(-a-i)-i\space Ei(i-a)]+\tan^{-1}(a)-\frac{\pi}{4}$$ In order to evaluate that, setting $a=0$, the tangent term vanishes, and we are left with the following: $$\frac{1}{2}[i \space Ei(-i)-i\space Ei(i)]=\frac{1}{2}[i\int_1^{\infty}\frac{e^{-t(-i)}}{t}dt-i\int_1^{\infty}\frac{e^{-t(i)}}{t}dt]$$ Getting both integrals together, and using $\frac{i}{2}=\frac{-1}{2i}$, we arrive at the following expression:$$-\int_{1}^{\infty}\frac{\frac{e^{it}-e{-it}}{2i}}{t}dt=-\int_1^{\infty}\frac{\sin(t)}{t}dt$$ which is the complementary sine integral evaluated at 1, and since the area under the positive real axis of that function is $\pi/2$, then the integral becomes $\pi/2-Si(1)$, where $Si(x)$ is the Sine integral.
And since $F(0)$ as we originally defined it $is$ the Sine integral evaluated at 1, then $c=\frac{\pi}{2}$.
I really hope this helps in some way... Could I ask what you needed to know the answer for? And if anyone has a better approach at evaluating these sort of integrals i'd be glad to know it.