I am thinking a bit stupidly but here comes a thought:

Let's find the indefinite integral of $\dfrac{e^x}{x}.$

....I know it's $\text{Ei}(x)+C$ but let's do it anyways

So firstly, I'm gonna let $\dfrac{1}{x}=t,$ leading us to $x=\dfrac{1}{t}.$

Differentiating gives $dx=-\dfrac{1}{t^2}dt,$ and therefore:

$\begin{aligned} \int \frac{e^x}{x}dx &= \int \frac{1}{x}\cdot e^x dx \\ & =\int t \cdot e^{\frac{1}{t}}\cdot\left(-\dfrac{1}{t^2}\right) dt \\ & =- \int \frac{e^{\frac{1}{t}}}{t} dt \\ & =- \int \frac{e^{\frac{1}{x}}}{x} dx \\ \end{aligned}$

From above, we get:

$\displaystyle \int \frac{e^x}{x} dx + \int \frac{e^{\frac{1}{x}}}{x} dx = 0 + (C)$

$\displaystyle \int \frac{e^x+e^{\frac{1}{x}}}{x} dx = 0 + (C)$

Differentiating both sides gives:

$\displaystyle \frac{e^x + e^{\frac{1}{x}}}{x} = 0$

....then...

$\displaystyle e^x+e^{\frac{1}{x}}=0$ ...?!

I didn't expect this to happen but I have this terrible migraine to think more deeply about this ._.

Yes I am stupid, sorry ;;>_>

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I am thinking a bit stupidly but here comes a thought:

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