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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.

- 3 months, 3 weeks ago

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

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

- 3 months, 3 weeks ago

You forgot about the arbitrary constant of integration.

- 4 months, 2 weeks ago

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,

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

- 4 months, 2 weeks ago

[ this comment is wrong ]

- 4 months, 2 weeks ago

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

- 4 months, 2 weeks ago

- 4 months, 2 weeks ago

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

- 4 months, 2 weeks ago

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.

- 4 months, 2 weeks ago

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

- 4 months, 2 weeks ago