- #1

pakmingki

- 93

- 1

make sure they are in the range of calculus 1-2 (anything before multivariable)

My teacher assigned some few hard integrals, and they are fun. I want to try moer.

thanks.

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- Thread starter pakmingki
- Start date

- #1

pakmingki

- 93

- 1

make sure they are in the range of calculus 1-2 (anything before multivariable)

My teacher assigned some few hard integrals, and they are fun. I want to try moer.

thanks.

- #2

yip

- 18

- 1

Try [tex] \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}} [/tex]

(forgot to put the integral sign in, it is now fixed)

(forgot to put the integral sign in, it is now fixed)

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- #3

itsjustme

- 44

- 0

sin(2x)cos(2x)dx

- #4

ObsessiveMathsFreak

- 406

- 8

[tex]\int e^{-x^2} dx[/tex]

- #5

VietDao29

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[tex]\int e^{-x^2} dx[/tex]

I doubt that it belongs to either Calculus 1 or Calculus 2 problems.

pakmingki said:... make sure they arein the range of calculus 1-2(anything before multivariable)...

- #6

Gib Z

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- #7

Kurdt

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One of my faves

- #8

pakmingki

- 93

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Ill give them a whirl sometime soon.

- #9

bob1182006

- 492

- 1

My favorite Integral so far is this:

[tex]\int \frac{dx}{(x^2+9)^3}[/tex]

It's general form is of

[tex]\int \frac{dx}{(x^2+a^2)^n}[/tex]

It has a really interesting answer

- #10

zoki85

- 1,187

- 227

Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

- #11

janhaa

- 97

- 4

Try this one...

[tex]\int \sqrt{\tan(x)}{\rm dx}[/tex]

[tex]\int \sqrt{\tan(x)}{\rm dx}[/tex]

- #12

bit188

- 45

- 0

That's a good one :rofl:Try this one...

[tex]\int \sqrt{\tan(x)}{\rm dx}[/tex]

- #13

Invictious

- 61

- 0

Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

Took me 5 minutes only

That question though, however, was just..simply amazing.

I suggest everyone try that question

- #14

Gib Z

Homework Helper

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I think the original poster has quite enough thanks...he hasn't actually done any of them yet.

- #15

JohnDuck

- 76

- 0

Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

I'm stumped but intrigued.

- #16

Equilibrium

- 82

- 0

[tex]\int \frac{1}{x^5+1}dx[/tex]

- #17

zoki85

- 1,187

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Took me 5 minutes only

That question though, however, was just..simply amazing.

I suggest everyone try that question

We are not all as clever as you Invictious :tongue:

- #18

DyslexicHobo

- 251

- 0

[tex]\int \frac{1}{x^5+1}dx[/tex]

I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

[tex]\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr[/tex]

I used u-substitution (well, r-substitution), where [tex]r = x^5 + 1[/tex]. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do [tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

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- #19

how can this even be integrated?:uhh:[tex]\int e^{-x^2} dx[/tex]

- #20

JohnDuck

- 76

- 0

how can this even be integrated?:uhh:

It can be proved that there's no elementary antiderivative, but you can use a trick from multivariable calculus involving a change to polar coordinates and the squeeze theorem to evaluate it. It's called a Gaussian integral.

Edit: Correction--the trick works for [tex]\int_{- \infty}^{\infty} e^{-x^{2}}dx[/tex]

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- #21

janhaa

- 97

- 4

how can this even be integrated?:uhh:

Observe that [tex]\,e^{-x^2}\,[/tex]is an even function, and we can integrate in two dimensions ):

[tex]I^2=(\int_{\mathbb{R}} e^{-x^2}){\rm dx})^2=(\int_{\mathbb{R}}e^{-x^2}{\rm dx})(\int_{\mathbb{R}}e^{-y^2}{\rm dy})[/tex]

[tex]I^2=\int_{\mathbb{R}} \int_{\mathbb{R}} e^{-(x^2+y^2)}{\rm dx}{\rm dy}[/tex]

Then change to polar coordinates:

[tex]I^2=\int_0^{2\pi}\int_0^{\infty} e^{-r^2} r {\rm dr} {\rm d\theta}=2\pi \int_0^{\infty} e^{-r^2} r {\rm dr}[/tex]

then substitution:

[tex]\, u = r^2 \,[/tex]

[tex]\frac{\rm du}{2r}={\rm dr}[/tex]

that is:

[tex]I^2=\pi \int_0^{\infty} e^{-u} {\rm du}= \pi[/tex]

finally:

[tex]I=\int_{\mathbb{R}} e^{-x^2} {\rm dx}=\int_{- \infty}^{\infty} e^{-x^2} {\rm dx}=\sqrt{\pi}[/tex]

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- #22

Office_Shredder

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Way to drop the ball on the limits at the end...

- #23

yip

- 18

- 1

I'm stumped but intrigued.

For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

[tex]\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr[/tex]

I used u-substitution (well, r-substitution), where [tex]r = x^5 + 1[/tex]. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do [tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

[tex] e^{-ix^{2}}=cosx^{2}-isinx^{2}[/tex]

[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx[/tex]

It is known that [tex]\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}[/tex]

[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}[/tex]

[tex]\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}[/tex]

[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)[/tex]

[tex]\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}[/tex]

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- #24

JohnDuck

- 76

- 0

I follow you up to here.For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:

[tex] e^{-ix^{2}}=cosx^{2}-isinx^{2}[/tex]

[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx[/tex]

It is known that [tex]\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}[/tex]

[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}[/tex]

[tex]\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}[/tex]

[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)[/tex]

Wha?[tex]\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}[/tex]

- #25

yip

- 18

- 1

Its simply taking the imaginary part of the integral, as the imaginary part of e^-ix^2 is -sinx^2

- #26

JohnDuck

- 76

- 0

Oh. That makes sense.

- #27

DyslexicHobo

- 251

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I followed you up to about the part where... uhh nevermind. Didn't catch any of that. :/

Way above my head. Thanks for the explanation, though. I don't even understand how we can even begin to integrate a transcendental function using limits of infinity. They don't have a value at infinity, so how can they be evaluated?

Way above my head. Thanks for the explanation, though. I don't even understand how we can even begin to integrate a transcendental function using limits of infinity. They don't have a value at infinity, so how can they be evaluated?

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- #28

JohnDuck

- 76

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[tex]\int_{a}^{\infty} f(x)dx[/tex]

are defined as such:

[tex]\lim_{b\rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{b\rightarrow \infty} F(b) - F(a)[/tex]

where F(x) is the antiderivative of f(x). If you don't understand how to take limits of functions at infinity, you should probably read up on limits again. Most calculus texts have a brief section on limits, and any introductory real analysis text certainly covers the topic thoroughly.

- #29

FlashStorm

- 17

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I am scared and frightened.

- #30

Kurdt

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I am scared and frightened.

:rofl::rofl:

- #31

DyslexicHobo

- 251

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[tex]\int_{a}^{\infty} f(x)dx[/tex]

are defined as such:

[tex]\lim_{b\rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{b\rightarrow \infty} F(b) - F(a)[/tex]

where F(x) is the antiderivative of f(x). If you don't understand how to take limits of functions at infinity, you should probably read up on limits again. Most calculus texts have a brief section on limits, and any introductory real analysis text certainly covers the topic thoroughly.

Sorry, I must have mis-worded myself. I meant that I don't understand how trigonometric functions can be evaluated as their inside approaches infinity. The function oscillates between 1 and -1, and never converges. I'm assuming that the Fundamental Theorem cannot be used here befause [tex]\lim_{x\rightarrow\infty} 2xcos(x^2)[/tex] (2xcos(x^2) is the anti-derivative of the starting function) cannot be evaluated, so it seems.

- #32

JohnDuck

- 76

- 0

Sorry, I must have mis-worded myself. I meant that I don't understand how trigonometric functions can be evaluated as their inside approaches infinity. The function oscillates between 1 and -1, and never converges. I'm assuming that the Fundamental Theorem cannot be used here befause [tex]\lim_{x\rightarrow\infty} 2xcos(x^2)[/tex] (2xcos(x^2) is the anti-derivative of the starting function) cannot be evaluated, so it seems.

That's actually not the antiderivative of [itex]\sin{x^{2}}[/itex]. You're right in that, for example, [itex]\lim_{x\rightarrow \infty} \sin{x}[/itex] doesn't converge in the real numbers. This can be proven pretty easily. However, depending on the function inside of sine, it may converge. Consider [itex]\lim_{x\rightarrow \infty} \sin{\frac{1}{x}}[/itex]--it converges to 0.

Edit: My gut says that [itex]\sin{x^{2}}[/itex] has no elementary antiderivative, but I'm not sure how one might prove or disprove this.

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- #33

Gib Z

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Well by integrating the Taylor series of sin(x^2) term by term we get:

[tex]\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{4n-1}}{(4n-1)(2n+1)!}[/tex]

which one could recognize as [tex]\sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)[/tex]

where [tex]S(u)=\int^u_0 \sin \left(\frac{1}{2} \pi x^2\right) dx[/tex].

As S(u) is not an elementary function, we have proved [itex]\sin x^2[/itex] has no elementary derivative.

[tex]\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{4n-1}}{(4n-1)(2n+1)!}[/tex]

which one could recognize as [tex]\sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)[/tex]

where [tex]S(u)=\int^u_0 \sin \left(\frac{1}{2} \pi x^2\right) dx[/tex].

As S(u) is not an elementary function, we have proved [itex]\sin x^2[/itex] has no elementary derivative.

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- #34

yip

- 18

- 1

Well by integrating the Taylor series of sin(x^2) term by term we get:

[tex]\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{4n-1}}{(4n-1)(2n+1)!}[/tex]

which one could recognize as [tex]\sqrt{\frac{\pi}{2}}S\left(\sqrt{\frac{2}{\pi}}x\right)[/tex]

where [tex]S(u)=\int^u_0 \sin \left(\frac{1}{2} \pi x^2\right) dx[/tex].

As S(u) is not an elementary function, we have proved [itex]\sin x^2[/itex] has no elementary anti-derivative.

Isn't that a sort of circular argument? If u assume S(u)=integral of sin(cx^2), where c=pi/2 does not have a closed form answer, aren't u also implicitly assuming that the integral of sin(x^2), the case where c=1, also does not have a closed form answer? I was under the impression that changing constants doesn't really affect integrability, only changing the variables that u are integrating with respect to would. I think that proving that something is not integratable would involve much more complicated arguments. This is just my opinion though. My attempt at saying something is not integrable would rely on the well known fact that e^-ix^2 has no closed form integral, and since the sum of the integral of isin(x^2) and cos(x^2) equals the integral of e^-ix^2, if a closed form integral did exist for sin(x^2), then one must also exist for cos(x^2) (since the cosine function is just a shifted sine function), which means a closed form integral exists for e^-ix^2, which is a contradiction.

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- #35

Gib Z

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