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anonymous

  • one year ago

Kinda tricky integral: \[\int_{-\infty}^\infty \frac{\cos x}{(x^2+1)(1+e^x)}\,dx\]

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  1. anonymous
    • one year ago
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    I've got some hints in mind, but I'm debating whether or not they give too much away :)

  2. myininaya
    • one year ago
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    *

  3. thomas5267
    • one year ago
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    I don't know but I kind of feel like it has something to do with Gaussian integral.

  4. anonymous
    • one year ago
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    It's related in that the result also contains \(\pi\), but beyond that I'm not sure... It'd be fascinating to see a derivation of this integral using the Guassian.

  5. thomas5267
    • one year ago
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    Convert to polar coordinate analogous to how you would evaluate Gaussian integral?

  6. thomas5267
    • one year ago
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    \[ f(x)=\frac{\cos x}{(x^2+1)(1+e^x)}\\ \begin{align*} f(x)f(y)&=\frac{\cos x}{(x^2+1)(1+e^x)}\frac{\cos y}{(y^2+1)(1+e^y)}\\ &=\frac{\cos(x)\cos(y)}{\left(r^2+x^2y^2+1)(1+e^x+e^y+e^{x+y}\right)},\,r^2=x^2+y^2 \end{align*} \] Aesthetically pleasing, probably of no use.

  7. anonymous
    • one year ago
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    Alright, here's my first hint: Notice that \(\dfrac{\cos x}{x^2+1}\) is even.

  8. anonymous
    • one year ago
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    Hint (1.a): \(y=-x\)

  9. ganeshie8
    • one year ago
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    Picking up on that hint : \[I=\int_{-\infty}^\infty \frac{\cos x}{(x^2+1)(1+e^x)}\,dx\] sub \(y=-x \) and get \[I=\int_{-\infty}^\infty \frac{e^x \cos x}{(x^2+1)(1+e^x)}\,dx\] Add them and get \[2I =\int_{-\infty}^\infty \frac{\cos x}{x^2+1}\,dx \]

  10. ganeshie8
    • one year ago
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    since the integrand is even, \[I =\int_{0}^\infty \frac{\cos x}{x^2+1}\,dx \]

  11. ganeshie8
    • one year ago
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    please do not give further hints, I think i got this :)

  12. anonymous
    • one year ago
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    So far so good! I had a lot of trouble making use of the "hidden" fact that \[\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}=1\] The rest of the solution can go any of several ways, I think. I'd love to see what you come up with.

  13. ganeshie8
    • one year ago
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    we may try DUT \[F(y) = \int_{0}^\infty \frac{\cos(xy)}{x^2+1}\,dx\] \[\implies F''(y) = -\int_{0}^\infty \frac{x^2\cos(xy)}{x^2+1}\,dx \] \[\implies -F''(y) + F(y) = \int_{0}^\infty \cos(xy)\,dx \]

  14. ganeshie8
    • one year ago
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    scratch that, doesn't converge

  15. anonymous
    • one year ago
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    Oooh, a related problem: \[\int_0^{\pi/2} \cos(\tan x)\,dx\](which is equivalent to the integral at hand, which you can see if you set \(t=\tan x\))

  16. anonymous
    • one year ago
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    @ganeshie8 I'm not at all surprised MSE has this one covered, but let's not spoil it for the rest. Based off of one of the answers I skimmed over, you seem to be on the right track with that ODE approach.

  17. ganeshie8
    • one year ago
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    I think its mostly done : \[F''(y) - F(y) = 0 \implies F(y) = C_1e^y + C_2e^{-y}\] finding \(C_1\) and \(C_2\) by choosing appropriate initial conditions should be a piece of cake

  18. IrishBoy123
    • one year ago
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    |dw:1442566006271:dw| do you not have to subtract them as the integration order needs to be reversed with the sub?

  19. ganeshie8
    • one year ago
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    differential also changes sign so that fixes the bounds..

  20. IrishBoy123
    • one year ago
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    duh :-)

  21. thomas5267
    • one year ago
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    The answer is \(\dfrac{\pi}{e}\)? Read something extremely similar to this on Wikipedia and want to confirm the answer before trying.

  22. thomas5267
    • one year ago
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    Did I missed by a factor of 2?

  23. anonymous
    • one year ago
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    Yep, the value of the integral is \(\dfrac{\pi}{2e}\), but it's not the answer that's important, it's how to get it :)

  24. thomas5267
    • one year ago
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    Wikipedia: Residue theorem. I don't know how Wikipedia did this but the answer is basically there. I don't even know line integral let alone the complex version of it.

  25. thomas5267
    • one year ago
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    Read at your own discretion!

  26. anonymous
    • one year ago
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    Right, contour integration is always an option, you just have to choose the right contour. My approach utilized the Laplace transform. First you parameterize the integral as follows. \[I(a)=\int_{-\infty}^\infty \frac{\cos ax}{(x^2+1)(1+e^x)}\,dx\] The change of variables/exponential identity reduces this to \[I(a)=\int_0^\infty \frac{\cos ax}{x^2+1}\,dx\] Take the Laplace transform: \[\mathcal{L}_s\{ I(a)\}=\int_0^\infty \left(\int_0^\infty \frac{\cos ax}{x^2+1}\,dx\right)e^{-as}\,da\] and some rearranging gives \[\mathcal{L}_s\{ I(a)\}=\int_0^\infty \left(\int_0^\infty e^{-as}\cos ax\,da\right)\frac{dx}{x^2+1}\] where the inner integral is just the Laplace transform of \(\cos ax\), i.e. \(\mathcal{L}_s\{\cos ax\}=\dfrac{s}{s^2+x^2}\). \[\mathcal{L}_s\{ I(a)\}=\int_0^\infty \frac{dx}{(x^2+1)(x^2+s^2)}\] Partial fractions: \[\frac{1}{(x^2+1)(x^2+s^2)}=\frac{1}{s^2-1}\left(\frac{1}{x^2+1}-\frac{1}{x^2+s^2}\right)\] And now we can integrate: \[\begin{align*}\mathcal{L}_s\{ I(a)\}&=\frac{1}{s^2-1}\int_0^\infty \left(\frac{1}{x^2+1}-\frac{1}{x^2+s^2}\right)\,dx\\[1ex] &=\frac{1}{s^2-1}\left(\frac{\pi}{2}-\frac{\pi}{2s}\right)\\[1ex] &=\frac{\pi}{2}\left(\frac{1}{s^2-1}-\frac{1}{s(s^2-1)}\right)\\[1ex] &=\frac{\pi}{2}\left(\frac{1}{s^2-1}+\frac{1}{s}-\frac{s}{s^2-1}\right) \end{align*}\] And finally, take the inverse transform. \[\begin{align*}\mathcal{L}_s\{ I(a)\}&=\frac{\pi}{2}\left(\frac{1}{s^2-1}+\frac{1}{s}-\frac{s}{s^2-1}\right)\\[1ex] I(a)&=\mathcal{L}_a^{-1}\{\cdots\}\\[1ex] &=\frac{\pi}{2}\left(\sinh a+1-\cosh a \right) \end{align*}\] Let \(a=1\) and we're done.

  27. IrishBoy123
    • one year ago
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    dark magic

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