Leibniz Integral Rule or differentiating under the integral sign.
Emily Wong 
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?
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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
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