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

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

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

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

Differentiating gives dx=1t2dt,dx=-\dfrac{1}{t^2}dt, and therefore:

exxdx=1xexdx=te1t(1t2)dt=e1ttdt=e1xxdx\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:

exxdx+e1xxdx=0+(C)\displaystyle \int \frac{e^x}{x} dx + \int \frac{e^{\frac{1}{x}}}{x} dx = 0 + (C)

ex+e1xxdx=0+(C)\displaystyle \int \frac{e^x+e^{\frac{1}{x}}}{x} dx = 0 + (C)

Differentiating both sides gives:

ex+e1xx=0\displaystyle \frac{e^x + e^{\frac{1}{x}}}{x} = 0

....then...

ex+e1x=0\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 Boi (보이)
2 years, 3 months ago

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1 vote

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e1/ttdte1/xxdx\int \frac{e^{1/t}}{t} dt \neq \int \frac{e^{1/x}}{x} dx. Remember x=1tx=\frac{1}{t}, and we are talking about indefinite integrals.

Abhishek Sinha - 2 years, 2 months ago

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

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

Boi (보이) - 2 years, 2 months ago

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You forgot about the arbitrary constant of integration.

Pi Han Goh - 2 years, 3 months ago

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Hmm...

then let exxdx+e1xxdx=C,\displaystyle \int \frac{e^x}{x}dx+\int \frac{e^{\frac{1}{x}}}{x}dx=C, but when you differentiate both sides,

ex+e1xx=0.\dfrac{e^x+e^{\frac{1}{x}}}{x}=0.

Boi (보이) - 2 years, 3 months ago

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[ this comment is wrong ]

Pi Han Goh - 2 years, 3 months 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

Boi (보이) - 2 years, 3 months ago

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@Boi (보이) Awhh damn it. I'll try to fix up my reply and answer your question shortly....

Pi Han Goh - 2 years, 3 months 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 - 2 years, 3 months ago

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@Pi Han Goh Suppose that we write F(x)  =  1xeuuduG(x)  =  1xe1uudu 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=u1v = u^{-1} gives F(x)  =  1x1e1vvdv  =  G(x1) F(x) \; = \; -\int_1^{x^{-1}} \frac{e^{\frac{1}{v}}}{v}\,dv \; = \; -G(x^{-1}) and hence F(x)+G(x1)  =  0 F(x) + G(x^{-1}) \; = \; 0 What you are saying is that F(x)+G(x)  =  c F(x) + G(x) \; = \; c which is not true. You are making the substitution t=x1t = x^{-1}, and then replacing tt by xx, instead of by x1x^{-1}. Since you are working with indefinite integrals, you cannot exchange the variables freely if they already have an interrelationship.

Mark Hennings - 2 years, 3 months ago

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

Boi (보이) - 2 years, 3 months ago

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