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 ;;>_>

#Calculus

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

You forgot about the arbitrary constant of integration.

Pi Han Goh - 2 years, 11 months 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,

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

Boi (보이) - 2 years, 11 months ago

[ this comment is wrong ]

Pi Han Goh - 2 years, 11 months ago

@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, 11 months ago

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

Pi Han Goh - 2 years, 11 months ago

@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, 11 months ago

@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 - 2 years, 11 months ago

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

Boi (보이) - 2 years, 11 months ago

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

Abhishek Sinha - 2 years, 10 months ago

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

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

Boi (보이) - 2 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$