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 ;;>_>
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H.M. 유
7 months, 2 weeks ago
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Top Newest\(\int \frac{e^{1/t}}{t} dt \neq \int \frac{e^{1/x}}{x} dx\). Remember \(x=\frac{1}{t}\), and we are talking about indefinite integrals.
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Yeah... Someone told me that in the below chain of comments and...
....I feel even dumber now. x'D
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You forgot about the arbitrary constant of integration.
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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,
\(\dfrac{e^x+e^{\frac{1}{x}}}{x}=0.\)
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[ this comment is wrong ]
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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
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Awhh damn it. I'll try to fix up my reply and answer your question shortly....
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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
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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.
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AHHH AH AH AH OKE THANKS now I understand, really, thank you!
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