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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 non-elementary functions. Thank you kindly.

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  1. freckles
    • one year ago
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    Could be indefinite or definite! Whateves!

  2. Kainui
    • one year ago
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    Here'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.

  3. freckles
    • one year ago
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    I think that one is a famous one. I think I remember seeing it done by taylor series.

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

  5. freckles
    • one year ago
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    or maybe I just need to play with that idea more

  6. freckles
    • one year ago
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    nevermind don't think it possible with taylor or power whatever you call it

  7. Kainui
    • one year ago
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    Haha yeah there's a nice way of doing it. =)

  8. Kainui
    • one year ago
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    The 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: \[\large \int\limits_{-\infty}^\infty e^{-x^2}dx = \sqrt{\int\limits_{-\infty}^\infty e^{-x^2}dx \int\limits_{-\infty}^\infty e^{-y^2}dy} \]

  9. freckles
    • one year ago
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    \[I=\int\limits_0^\infty e^{-x^2} dx \\ I \cdot I=\int\limits_0^\infty e^{-x^2} dx \cdot \int\limits_0^\infty e^{-x^2} dx \\ \text{ change up the variable on one } \\ I^2=\int\limits_0^\infty e^{-x^2} dx \cdot \int\limits_0^\infty e^{-y^2} dy \\ I^2=\int\limits_0^\infty \int\limits_0^\infty e^{-x^2} e^{-y^2} dx dy\] I don't know why the last two lines are equivalent?

  10. Kainui
    • one year ago
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    The 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.

  11. freckles
    • one year ago
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    yes that is actually

  12. freckles
    • one year ago
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    \[I^2 =\int\limits_0^\infty \int\limits_0^\infty e^{-(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits_0^\infty \int\limits_0^\infty r e^{-r^2} dx dy \\ 0<x<\infty \\ 0<y<\infty \] 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 \[\int\limits_{- \infty}^\infty e^{-x^2} dx=2 \cdot I=2 \cdot \frac{\sqrt{\pi}}{2}=\sqrt{\pi}\]

  13. Kainui
    • one year ago
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    Awesome, alright I guess I need to find something harder to send your way.

  14. freckles
    • one year ago
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    lol when you say harder I know it will be because you are crazy good at math

  15. freckles
    • one year ago
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    by the way I did make a mistake in one of my lines

  16. freckles
    • one year ago
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    if 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

  17. freckles
    • one year ago
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    \[I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{-(x^2+y^2)} dx dy \\ r^2=x^2+y^2 \\ \\ I^2 =\int\limits\limits_0^\infty \int\limits\limits_0^\infty e^{-r^2} dx dy \\ 0<x<\infty \\ 0<y<\infty \\ I^2=\int\limits\limits_0^\frac{\pi}{2} \int\limits\limits_0^\infty r e^{-r^2} dr d \theta \\ I^2=\int\limits\limits_0^\frac{\pi}{2}[\frac{-1}{2}e^{-r^2}]_0^{ z \rightarrow \infty}] d \theta \\ I^2=\int\limits\limits_0^\frac{\pi}{2}[\frac{-1}{2}e^{-z^2}+\frac{1}{2}] d \theta \\ I^2 = \int\limits\limits_0^\frac{\pi}{2}\frac{1}{2} d \theta \\ I^2=\frac{1}{2} \frac{\pi}{2} \\ I=\frac{\sqrt{\pi}}{2} \]

  18. Kainui
    • one year ago
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    Hmmm 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%20-infty%20to%20infty%20e%5E-x%5E2dx&t=crmtb01

  19. freckles
    • one year ago
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    yeah that is the same I got I'm just saying I put the r in too early

  20. Kainui
    • one year ago
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    Ohhh ok I see. Alright here's a fun one haha: \[\large \int\limits_0^\infty \frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx \text{ when } a>0\]

  21. freckles
    • one year ago
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    now this one I do believe is entirely new to me let me think a bit

  22. freckles
    • one year ago
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    so 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?

  23. freckles
    • one year ago
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    you know what nevermind it is probably some kinda pattern or whatever

  24. Kainui
    • one year ago
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    Yep, 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: \[\large \int\limits_0^\infty \lfloor x \rfloor e^{-x} dx\]

  25. freckles
    • one year ago
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    hey @xapproachesinfinity we are just having integral fun if you want to join in in either trying to answer or give more questions

  26. Kainui
    • one year ago
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    Here'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 non-negative integer \(k\). This one originally comes from here: http://openstudy.com/study#/updates/54bc2960e4b0c3c33929e555

  27. xapproachesinfinity
    • one year ago
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    integral party lol i have seen that first integral many times ant its value never question how it is root(pi)

  28. freckles
    • one year ago
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    @Kainui is there any more ways to integrate the first one you mentioned besides the user of "polar-ness" That might be easiest way right? Like the taylor/power series is a complete dead end?

  29. freckles
    • one year ago
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    and wow you guys did put a lot of work into that one problem

  30. freckles
    • one year ago
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    http://openstudy.com/users/sithsandgiggles#/updates/54a1af22e4b054f0c3b34fcf I never understood this parameter thing

  31. anonymous
    • one year ago
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    Hi, please accept this monstrosity: http://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1

  32. Kainui
    • one year ago
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    There'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!

  33. freckles
    • one year ago
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    like 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

  34. freckles
    • one year ago
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    @sithsandgiggles has asked a lot of integral questions

  35. freckles
    • one year ago
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    hard ones too

  36. freckles
    • one year ago
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    \[\large \int\limits\limits_0^\infty \frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx \text{ when } a>0 \\ f(t)=\tan^{-1}(tx) \\ f(a)=\tan^{-1}(ax) \\ f(1)=\tan^{-1}(1x)=\tan^{-1}(x) \\ \int\limits_0^\infty \frac{f(a)-f(1)}{x} dx \\ \int\limits_0^\infty \frac{1}{x} \int\limits_1^a f'(t) dt\] trying to play with this

  37. freckles
    • one year ago
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    though I don't think that is exactly what I want

  38. xapproachesinfinity
    • one year ago
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    just one thing the first one how likely a person would think they need to change the dummy variable isn't that somewhat unlikely?

  39. freckles
    • one year ago
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    guess it kind of depends on if one is familiar with polar-need integration if that is the case the odds that person does that goes a little up

  40. freckles
    • one year ago
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    polar-ness*

  41. freckles
    • one year ago
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    you know like since we know x^2+y^2=r^2

  42. xapproachesinfinity
    • one year ago
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    i agree

  43. freckles
    • one year ago
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    then eventually using the r dr dtheta= dx dy thing so the integration actually becomes doable

  44. freckles
    • one year ago
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    \[\int\limits f'(tx) dt \\ \text{ Let } u(t)=tx \\ \frac{d(u(t))}{dt}=\frac{d(tx)}{dt} \\ \frac{d(u(t))}{dt}=x \\ \frac{du}{dt}=x \\ du=x dt \\ \frac{1}{x} du =dt \\ \int\limits f'(tx) dx=\frac{1}{x} \int\limits f'(u) du=\frac{1}{x}( f(u)+C)\] 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

  45. freckles
    • one year ago
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    I guess we can try the fubini theorem thingy

  46. freckles
    • one year ago
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    \[\int\limits_1^a \int\limits_0^\infty f'(tx) dx dt \\ \text{ Let } u(x)=tx \\ u'(x)=t \\ du=t dx \\ \frac{1}{t} du =dx \\ \int\limits_1^a \int\limits_0^\infty \frac{1}{t} f'(u) du dt \\ \int\limits_1^a \frac{1}{t } \int\limits_0^\infty f'(u) du dt \\ \int\limits_1^a \frac{1}{t} f(u)|_0^{z \rightarrow \infty} dt \\ \] 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

  47. Kainui
    • one year ago
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    Another favorite: \[\large \int\limits_0^\infty x^n e^{-x}dx\]

  48. xapproachesinfinity
    • one year ago
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    i did IBP for the last one \[I_k=n(n-1)(n-2).....(n-k)\int\limits_0^\infty x^{n-k-1}e^{-x}dx \]

  49. xapproachesinfinity
    • one year ago
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    may be on the wrong truck

  50. Kainui
    • one year ago
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    I 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

  51. Kainui
    • one year ago
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    I 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 ;)

  52. xapproachesinfinity
    • one year ago
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    hmm there is an easier way

  53. xapproachesinfinity
    • one year ago
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    how about we look I1, I2..... and generalize but that is gonna be by parts as well

  54. xapproachesinfinity
    • one year ago
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    anyways have a good night rest :)

  55. xapproachesinfinity
    • one year ago
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    it is definitely n! that last one

  56. xapproachesinfinity
    • one year ago
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    doing it with cases give you the idea of the pattern that is raising

  57. xapproachesinfinity
    • one year ago
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    somehow for the second i got \[I=\frac{\pi}{2}(1-a)\] for a>0

  58. anonymous
    • one year ago
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    A 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.

  59. xapproachesinfinity
    • one year ago
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    for the second i send x to other side to to get I'=xI not sure of this is valid lol

  60. xapproachesinfinity
    • one year ago
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    then i manipulate the integral with arctan ax and artan x only

  61. xapproachesinfinity
    • one year ago
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    does that sound doable?

  62. freckles
    • one year ago
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    what did you do for second one?

  63. freckles
    • one year ago
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    i'm still lost on the second one

  64. xapproachesinfinity
    • one year ago
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    \[I=\int \frac{\arctan ax-\arctan x}{x}dx\] \[xI=\int \arctan ax-\arctan xdx\] try to do that integrand and then divide by x later

  65. xapproachesinfinity
    • one year ago
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    oh no i just realized a mistake

  66. xapproachesinfinity
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    that was ridulucous of me lol

  67. xapproachesinfinity
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    perhaps Taylor would work

  68. xapproachesinfinity
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    gotta go :) take care

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

  70. freckles
    • one year ago
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    and of course you already said a>0

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