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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:

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

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 ;;>_>

Note by H.M. 유
4 months, 2 weeks ago

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\(\int \frac{e^{1/t}}{t} dt \neq \int \frac{e^{1/x}}{x} dx\). Remember \(x=\frac{1}{t}\), and we are talking about indefinite integrals.

Abhishek Sinha - 3 months, 3 weeks ago

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Yeah... Someone told me that in the below chain of comments and...

....I feel even dumber now. x'D

H.M. 유 - 3 months, 3 weeks ago

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You forgot about the arbitrary constant of integration.

Pi Han Goh - 4 months, 2 weeks ago

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Hmm...

then let \(\displaystyle \int \frac{e^x}{x}dx+\int \frac{e^{\frac{1}{x}}}{x}dx=C,\) but when you differentiate both sides,

\(\dfrac{e^x+e^{\frac{1}{x}}}{x}=0.\)

H.M. 유 - 4 months, 2 weeks ago

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[ this comment is wrong ]

Pi Han Goh - 4 months, 2 weeks ago

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@Pi Han Goh Yes

The integral of 1 - cos^2 x is

The integral of 1 - the integral of cos^2 x

= x - the integral of cos^2 x + C

Therefore

The integral of sin^2 x + the integral of cos^2 x = x + C

Differentiate and we get

sin^2 x + cos^2 x = 1

H.M. 유 - 4 months, 2 weeks ago

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@H.M. 유 Awhh damn it. I'll try to fix up my reply and answer your question shortly....

Pi Han Goh - 4 months, 2 weeks ago

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@Pi Han Goh Arghhh, curses! I can't properly explain it. I think got something to do with inverse functions and 1-1 or something.

Summoning the great @Mark Hennings

Pi Han Goh - 4 months, 2 weeks ago

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@Pi Han Goh Suppose that we write \[ F(x) \; = \; \int_1^x \frac{e^u}{u}\,du \hspace{2cm} G(x) \; = \; \int_1^x \frac{e^{\frac{1}{u}}}{u}\,du \] Then the substitution \(v = u^{-1}\) gives \[ F(x) \; = \; -\int_1^{x^{-1}} \frac{e^{\frac{1}{v}}}{v}\,dv \; = \; -G(x^{-1}) \] and hence \[ F(x) + G(x^{-1}) \; = \; 0 \] What you are saying is that \[ F(x) + G(x) \; = \; c \] which is not true. You are making the substitution \(t = x^{-1}\), and then replacing \(t\) by \(x\), instead of by \(x^{-1}\). Since you are working with indefinite integrals, you cannot exchange the variables freely if they already have an interrelationship.

Mark Hennings - 4 months, 2 weeks ago

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@Mark Hennings AHHH AH AH AH OKE THANKS now I understand, really, thank you!

H.M. 유 - 4 months, 2 weeks ago

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