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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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    $\begingroup$ I can't see any work to tell you what you did wrong. And to make bigger integrals/math in general, use double dollar signs: $$....$$ $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 18:14
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    $\begingroup$ The step you're at (integrating $\int e^{-ax}\sin(x)dx$) can be solved with two integration by parts. See here $\endgroup$ – Mark Feb 18 '17 at 18:15
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    $\begingroup$ Additionally, you can rewrite it as $\int e^{-ax}\Re(e^{ix})dx$, integrate it to get $\Re(\frac{1}{i-a} e^{(i-a)x})$, then simplify. $\endgroup$ – Mark Feb 18 '17 at 18:17
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    $\begingroup$ @smaude It is somewhat more complex analysis type of method. Note that $i=\sqrt{-1}$ in his comment. It follows from what is known as Euler's formula. $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 18:24
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    $\begingroup$ @SimplyBeautifulArt meh, I disagree. I think comments can be used to get an initial feel for what the OP may or may not know, and is a good place to throw out some initial ideas that you have to solve a problem (especially lengthy ones) without drawing out every step for the OP, especially when you don't know if the ideas will work. $\endgroup$ – Brevan Ellefsen Feb 18 '17 at 18:25
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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.

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    $\begingroup$ The original problem isn't solved. The integral you started with is the partial derivative of something. $\endgroup$ – Rafa Budría Feb 18 '17 at 18:42
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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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    $\begingroup$ Thanks this is very helpful. And what I needed the answer for? Just curiosity I guess. I was trying some new methods of integration and when I came across this I couldn't solve it. $\endgroup$ – smaude Feb 18 '17 at 22:13
  • $\begingroup$ @smaude Bleh, not all integrals were born to be solved you know. :-( $\endgroup$ – Simply Beautiful Art Feb 20 '17 at 0:45
  • $\begingroup$ @SimplyBeautifulArt I think thats what makes them so interesting though! $\endgroup$ – Fede Poncio Feb 20 '17 at 2:38

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