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freckles
 one year ago
Integral Jungle! Are there any integrals you know of that seem like really hard to do but you just need to find a cute way to manipulate things? Oh and also please none that involve nonelementary functions. Thank you kindly.
freckles
 one year ago
Integral Jungle! Are there any integrals you know of that seem like really hard to do but you just need to find a cute way to manipulate things? Oh and also please none that involve nonelementary functions. Thank you kindly.

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freckles
 one year ago
Best ResponseYou've already chosen the best response.4Could be indefinite or definite! Whateves!

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Here's a good one if you've never seen this before: \[\large \int\limits_{\infty}^\infty e^{x^2}dx\] I know a handful of other fun integrals too that I really enjoy.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I think that one is a famous one. I think I remember seeing it done by taylor series.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[e^x=\sum_{n=0}^\infty \frac{x^n}{n!} \\ e^{x^2}=\sum_{n=0}^\infty \frac{(x^2)^n}{n!}=\sum_{n=0}^\infty \frac{(1)^nx^{2n}}{n!}\] \[\int\limits_0^\infty e^{x^2} dx \\ e^{x^2} \approx 1x^2+\frac{x^4}{2}\frac{x^6}{6}+ \cdots\] hmm maybe my memory isn't right

freckles
 one year ago
Best ResponseYou've already chosen the best response.4or maybe I just need to play with that idea more

freckles
 one year ago
Best ResponseYou've already chosen the best response.4nevermind don't think it possible with taylor or power whatever you call it

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Haha yeah there's a nice way of doing it. =)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3The trick is sorta weird if you've never seen it before. The first step is to square the integral, but you need to rewrite the other integral in terms of a different dummy variable. Since it's squared, you take the square root, so these integrals are exactly the same but the one on the right lets us play around and combine them into a double integral: \[\large \int\limits_{\infty}^\infty e^{x^2}dx = \sqrt{\int\limits_{\infty}^\infty e^{x^2}dx \int\limits_{\infty}^\infty e^{y^2}dy} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[I=\int\limits_0^\infty e^{x^2} dx \\ I \cdot I=\int\limits_0^\infty e^{x^2} dx \cdot \int\limits_0^\infty e^{x^2} dx \\ \text{ change up the variable on one } \\ I^2=\int\limits_0^\infty e^{x^2} dx \cdot \int\limits_0^\infty e^{y^2} dy \\ I^2=\int\limits_0^\infty \int\limits_0^\infty e^{x^2} e^{y^2} dx dy\] I don't know why the last two lines are equivalent?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3The reason we can combine them together is because they're summed over a different independent variables so the multiplication basically is just distributing. For example: \[\large \left( \sum_{n=1}^3 a_n \right)\left( \sum_{m=1}^3 b_m \right) = \sum_{n=1}^3 \sum_{m=1}^3 a_n b_m \] \[(a_1+a_2+a_3)(b_1+b_2+b_3) = a_1b_1+a_1b_2+a_1b_3+ \cdots\] I don't know if that's convincing enough or not.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[I^2 =\int\limits_0^\infty \int\limits_0^\infty e^{(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits_0^\infty \int\limits_0^\infty r e^{r^2} dx dy \\ 0<x<\infty \\ 0<y<\infty \] That is everything in first quadrant so that means \[0<r<\infty \\ 0<\theta<\frac{\pi}{2}\] so we have \[I^2=\int\limits_0^\frac{\pi}{2} \int\limits_0^\infty r e^{r^2} dr d \theta \\ I^2=\int\limits_0^\frac{\pi}{2}[\frac{1}{2}e^{r^2}]_0^{ z \rightarrow \infty}] d \theta \\ I^2=\int\limits_0^\frac{\pi}{2}[\frac{1}{2}e^{z^2}+\frac{1}{2}] d \theta \\ I^2 = \int\limits_0^\frac{\pi}{2}\frac{1}{2} d \theta \\ I^2=\frac{1}{2} \frac{\pi}{2} \\ I=\frac{\sqrt{\pi}}{2}\] So \[\int\limits_{ \infty}^\infty e^{x^2} dx=2 \cdot I=2 \cdot \frac{\sqrt{\pi}}{2}=\sqrt{\pi}\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Awesome, alright I guess I need to find something harder to send your way.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4lol when you say harder I know it will be because you are crazy good at math

freckles
 one year ago
Best ResponseYou've already chosen the best response.4by the way I did make a mistake in one of my lines

freckles
 one year ago
Best ResponseYou've already chosen the best response.4if I was going to leave the dx dy there while I figured out the limits then that one r shouldn't haven been there yet since dx dy=r dr d theta

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{r^2} dx dy \\ 0<x<\infty \\ 0<y<\infty \\ I^2=\int\limits\limits_0^\frac{\pi}{2} \int\limits\limits_0^\infty r e^{r^2} dr d \theta \\ I^2=\int\limits\limits_0^\frac{\pi}{2}[\frac{1}{2}e^{r^2}]_0^{ z \rightarrow \infty}] d \theta \\ I^2=\int\limits\limits_0^\frac{\pi}{2}[\frac{1}{2}e^{z^2}+\frac{1}{2}] d \theta \\ I^2 = \int\limits\limits_0^\frac{\pi}{2}\frac{1}{2} d \theta \\ I^2=\frac{1}{2} \frac{\pi}{2} \\ I=\frac{\sqrt{\pi}}{2} \]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Hmmm well you seem to have messed up twice or something along the lines so it cancelled out or something? http://www.wolframalpha.com/input/?i=integral%20infty%20to%20infty%20e%5Ex%5E2dx&t=crmtb01

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yeah that is the same I got I'm just saying I put the r in too early

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Ohhh ok I see. Alright here's a fun one haha: \[\large \int\limits_0^\infty \frac{\tan^{1}(ax)\tan^{1}(x)}{x}dx \text{ when } a>0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4now this one I do believe is entirely new to me let me think a bit

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so that means a can be 1 and if a is 1 then the integrand is 0 which mean the output is 0 for a=1 so should the inequality include 1?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you know what nevermind it is probably some kinda pattern or whatever

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Yep, you might have to be careful about what a is when you're defining your answer, like specify some things perhaps. I don't have the answer with me but it's not too difficult if you know the trick =P Here's another good one: \[\large \int\limits_0^\infty \lfloor x \rfloor e^{x} dx\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4hey @xapproachesinfinity we are just having integral fun if you want to join in in either trying to answer or give more questions

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Here's one that took several days to figure out, but we did it: Integrate: \[\int_0^\infty \frac{(\sin x)^{2k+1}}{x}\,dx\] for nonnegative integer \(k\). This one originally comes from here: http://openstudy.com/study#/updates/54bc2960e4b0c3c33929e555

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0integral party lol i have seen that first integral many times ant its value never question how it is root(pi)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4@Kainui is there any more ways to integrate the first one you mentioned besides the user of "polarness" That might be easiest way right? Like the taylor/power series is a complete dead end?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4and wow you guys did put a lot of work into that one problem

freckles
 one year ago
Best ResponseYou've already chosen the best response.4http://openstudy.com/users/sithsandgiggles#/updates/54a1af22e4b054f0c3b34fcf I never understood this parameter thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hi, please accept this monstrosity: http://math.stackexchange.com/questions/562694/integralint11frac1xsqrtfrac1x1xlnleftfrac2x22x1

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3There's one integral that's really fun to do with integration by parts, but you can also do it with linear algebra, this one: \[\large \int\limits e^{ax}\cos(bx)dx\] The trick is pretty fun and you can use a similar trick for solving some differential equations too: Find basis vectors: \[y_1 = e^{ax}\sin(bx) \\ y_2 = e^{ax}\cos(bx)\] The derivative of these functions is a linear combination of these functions so now instead of doing integration by parts, integration is just inverting a 2x2 matrix!

freckles
 one year ago
Best ResponseYou've already chosen the best response.4like why does \[I(b)=\int\limits_0^\infty \frac{\ln(bx)}{1+x^2} dx =\int\limits_0^\infty \frac{\ln(b)+\ln(x)}{1+x^2} dx \\ =\int\limits_0^\infty \frac{\ln(b)}{1+x^2} dx+\int\limits_0^\infty \frac{\ln(x)}{1+x^2} dx \\ =\ln(b) \int\limits_0^\infty \frac{1}{1+x^2} dx+\int\limits_0^\infty \frac{\ln(x)}{1+x^2} dx \\ I'(b)=\frac{1}{b} \int\limits_0^\infty \frac{1}{1+x^2}dx+0\] oops nevermind I get it

freckles
 one year ago
Best ResponseYou've already chosen the best response.4@sithsandgiggles has asked a lot of integral questions

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\large \int\limits\limits_0^\infty \frac{\tan^{1}(ax)\tan^{1}(x)}{x}dx \text{ when } a>0 \\ f(t)=\tan^{1}(tx) \\ f(a)=\tan^{1}(ax) \\ f(1)=\tan^{1}(1x)=\tan^{1}(x) \\ \int\limits_0^\infty \frac{f(a)f(1)}{x} dx \\ \int\limits_0^\infty \frac{1}{x} \int\limits_1^a f'(t) dt\] trying to play with this

freckles
 one year ago
Best ResponseYou've already chosen the best response.4though I don't think that is exactly what I want

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0just one thing the first one how likely a person would think they need to change the dummy variable isn't that somewhat unlikely?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4guess it kind of depends on if one is familiar with polarneed integration if that is the case the odds that person does that goes a little up

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you know like since we know x^2+y^2=r^2

freckles
 one year ago
Best ResponseYou've already chosen the best response.4then eventually using the r dr dtheta= dx dy thing so the integration actually becomes doable

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\int\limits f'(tx) dt \\ \text{ Let } u(t)=tx \\ \frac{d(u(t))}{dt}=\frac{d(tx)}{dt} \\ \frac{d(u(t))}{dt}=x \\ \frac{du}{dt}=x \\ du=x dt \\ \frac{1}{x} du =dt \\ \int\limits f'(tx) dx=\frac{1}{x} \int\limits f'(u) du=\frac{1}{x}( f(u)+C)\] so maybe I can redefined by f from above so we have \[f'(tx)=\frac{\arctan(tx)}{x} \\ f'(ax)=\frac{\arctan(ax)}{x} \\ f'(1x)=\frac{\arctan(1x)}{x} \\ f'(ax)f'(1x)=\frac{\arctan(ax)}{x}\frac{\arctan(1x)}{x}=\frac{\arctan(ax)\arctan(x)}{x} \\ \\ \text{ so we could say } \\ \int\limits_0^\infty \frac{\arctan(ax)\arctan(x)}{x} dx \\ \int\limits_0^\infty \int\limits_1^a f'(tx) dt dx\] I don't know if this helps

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I guess we can try the fubini theorem thingy

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\int\limits_1^a \int\limits_0^\infty f'(tx) dx dt \\ \text{ Let } u(x)=tx \\ u'(x)=t \\ du=t dx \\ \frac{1}{t} du =dx \\ \int\limits_1^a \int\limits_0^\infty \frac{1}{t} f'(u) du dt \\ \int\limits_1^a \frac{1}{t } \int\limits_0^\infty f'(u) du dt \\ \int\limits_1^a \frac{1}{t} f(u)_0^{z \rightarrow \infty} dt \\ \] what is f(u)... \[tx=u \\ \text{ so I think } f'(u)=\frac{\arctan(u)}{\frac{u}{t}}= t \frac{\arctan(u)}{u} \\ \text{ so I need \to evaluate } \\ \int\limits \frac{\arctan(u)}{u} du \\ y=\arctan(u) \\ \tan(y)=u \\ \sec^2(y) dy=du \\ \int\limits \frac{y}{\tan(y)} \sec^2(y) dy \\ \int\limits y \frac{\cos(y)}{\sin(y)} \frac{1}{\cos^2(y)} dy \\ \int\limits y \frac{1}{\sin(y) \cos(y)} dy \\ \int\limits 2y \csc(2y) d y \\ v=2y \\ dv =2 dy \\ \frac{1}{2} dv =dy \\ \frac{1}{2} \int\limits v \csc(v) dv\] maybe I can't integrate that

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Another favorite: \[\large \int\limits_0^\infty x^n e^{x}dx\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i did IBP for the last one \[I_k=n(n1)(n2).....(nk)\int\limits_0^\infty x^{nk1}e^{x}dx \]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0may be on the wrong truck

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3I woke up really early so I'm sorta tired. I'll let ya think about these and play around, I'll keep coming back tomorrow and give hints and more integrals of course if you like =P

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3I think you're on the right path for that last one @xapproachesinfinity and I think that will work, however there's another tricky way to do it too ;)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm there is an easier way

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0how about we look I1, I2..... and generalize but that is gonna be by parts as well

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0anyways have a good night rest :)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0it is definitely n! that last one

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0doing it with cases give you the idea of the pattern that is raising

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0somehow for the second i got \[I=\frac{\pi}{2}(1a)\] for a>0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A recent favorite approach of mine: There a lot of integrals you can evaluate using the Laplace transform. Here's a good "model": \[\int_0^\infty \frac{\sin^3 x}{x^2+1}\,dx\] for natural \(k\). It involves introducing a parameter, as you would when differentiating under the integral sign, but there's no differentiation involved. The hardest part is probably just reducing the numerator into "easier" trig functions via identities. The tedious part is taking the forward/inverse transforms to arrive at the result.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0for the second i send x to other side to to get I'=xI not sure of this is valid lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0then i manipulate the integral with arctan ax and artan x only

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0does that sound doable?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4what did you do for second one?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4i'm still lost on the second one

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0\[I=\int \frac{\arctan ax\arctan x}{x}dx\] \[xI=\int \arctan ax\arctan xdx\] try to do that integrand and then divide by x later

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0oh no i just realized a mistake

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0that was ridulucous of me lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0perhaps Taylor would work

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0gotta go :) take care

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\int\limits f dx=F+C \\ \text{ where } F'=f \\ \text{ we have } \\ \int\limits_0^\infty \frac{\arctan(ax)\arctan(x)}{x} dx \\ f(tx)=\arctan(tx) \\ \int\limits_0^\infty \frac{1}{x}[\arctan(tx)]_{t=1}^{t=a} dx \\ \\ \text{ Let } f(tx)=\arctan(tx) \\ \int\limits_0^\infty \frac{1}{x}[f(tx)]_{t=1}^{t=a} dx \\ \int\limits_0^\infty [\frac{f(tx)}{x}]_{t=1}^{t=a} dx \\ g(t)=\frac{f(tx)}{x} \\ \frac{dg}{dt}=\frac{1}{x} \cdot x \cdot \frac{df(tx)}{dt}=\frac{df(tx)}{dt} \\ \\ \] \[\int\limits_0^\infty \int\limits_1^a \frac{dg}{dt} dt dx \\ \int\limits_1^a \int\limits_0^\infty \frac{dg}{dt} dx dt \\ \] \[\int\limits_1^a [\frac{1}{t} f(tx)]_{x=0}^{x=z \rightarrow \infty } dt \\ \int\limits_1^a[\frac{1}{t}[f(t \cdot z)f(t \cdot 0)] dt \\ \int\limits_1^a \frac{1}{t}[\arctan(tz)\arctan(0)] dt \\ \int\limits_1^a \frac{1}{t}[\frac{\pi}{2}0] dt \\ \ln(t)_1^a \cdot \frac{\pi}{2} \\ [\ln(a)\ln(1)] \frac{\pi}{2} \\ \frac{\ln(a) \pi}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4and of course you already said a>0
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