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I'm trying to compute $\int \frac{e^{-x}\sin x}{x} \ dx$. This resembles integrating $\int e^{-x}\sin x \ dx$ so my main thought about how to approach this is by finding some connection with that. I know that $\int e^{-x}\sin x \ dx = -\frac 1 2 e^{-x}(\sin x + \cos x)$. If I try to use integration by parts setting

$$u=x^{-1}$$

and

$$dv=e^{-x}\sin x \ dx$$

then we get

$$\frac{du}{dx}=-\frac{1}{x^2}$$

and also

$$v=-\frac 1 2 e^{-x}(\sin x + \cos x)$$

Then the integral becomes

$$ \int u \ dv = uv-\int v\ du = x^{-1}\left(-\frac 1 2 e^{-x}(\sin x + \cos x) \right) - \int \left( -\frac 1 2 e^{-x}(\sin x + \cos x) \right) \left( -\frac{1}{x^2} \right) \ dx $$

I seem to have made things worse. I am sorry.

If I try to make any other assignment of $u$ and $dv$ in integration by parts, I don't foresee them going any better. No substitution seems helpful, either $u$ or trig. I don't think I can integrate twice and get back to where I started. That almost seems possible if I take what I have done above and now choose $\frac 1 {x^2}$ for integration, except that the derivative part will now have much more in it than I started with. I think my bag of tricks may be empty.

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    $\begingroup$ The antiderivative can be expressed in terms of a special function, the exponential integral. And no need to apologize :) $\endgroup$
    – user170231
    Feb 26 at 22:13
  • $\begingroup$ Have you tried putting it into Wolfram Alpha? $\endgroup$
    – Some Guy
    Feb 26 at 22:14
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    $\begingroup$ Clearly, the integral is $\int_{0}^{x}\frac{e^{-t}\sin t}{t} \, dt$ + C ;) $\endgroup$
    – Joe
    Feb 26 at 22:15
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    $\begingroup$ Yes but how @vitamind, anyone can put this integral into Wolfam and get an answer, but how $\endgroup$
    – Some Guy
    Feb 26 at 22:18
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    $\begingroup$ @SomeGuy That's why it's a comment. To answer your question: Use Eulers formula. $\endgroup$
    – vitamin d
    Feb 26 at 22:19
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Since no one is going to post an answer I will prove that $$I = \int \frac{e^{-x}\sin x}{x}\,\mathrm{d}x = -\frac{\mathrm{i}\left(\operatorname{Ei}\left(\left(\mathrm{i}-1\right)x\right)-\operatorname{Ei}\left(-\left(\mathrm{i}+1\right)x\right)\right)}{2}+C,$$ where $\operatorname{Ei}$ is the expontial integral and $C$ is an arbitrary real constant.

Lemma. Let $z$ be a complex number. Then $$\sin z = \frac{e^{iz}-e^{-iz}}{2i}.$$ If we rewrite the sine in the integral, we may add exponents and use linearity of the integral to get to $$I = \frac{1}{2i}\int \frac{e^{ix-x}}{x}\,\mathrm{d}x-\frac{1}{2i}\int \frac{e^{-ix-x}}{x}\,\mathrm{d}x.$$ Substitute in the first integral $u_1=(i-1)x$ and in the second one $u_2=-(i+1)x$. $$\frac{1}{2i}\operatorname{Ei}(u_1)-\frac{1}{2i}\operatorname{Ei}(u_2).$$ Substituting $u_{1,2}$ back yields $$\frac{\operatorname{Ei}((i-1)x)-\operatorname{Ei}(-(i-1)x)}{2i}+C.$$ The last step is a matter of taste. You can write $\tfrac{1}{i}$ as $-i$, so $$\frac{-i(\operatorname{Ei}((i-1)x)-\operatorname{Ei}(-(i-1)x))}{2}+C.$$

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