anonymous
  • anonymous
Kinda tricky integral: \[\int_{-\infty}^\infty \frac{\cos x}{(x^2+1)(1+e^x)}\,dx\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I've got some hints in mind, but I'm debating whether or not they give too much away :)
myininaya
  • myininaya
*
thomas5267
  • thomas5267
I don't know but I kind of feel like it has something to do with Gaussian integral.

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anonymous
  • anonymous
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.
thomas5267
  • thomas5267
Convert to polar coordinate analogous to how you would evaluate Gaussian integral?
thomas5267
  • thomas5267
\[ f(x)=\frac{\cos x}{(x^2+1)(1+e^x)}\\ \begin{align*} f(x)f(y)&=\frac{\cos x}{(x^2+1)(1+e^x)}\frac{\cos y}{(y^2+1)(1+e^y)}\\ &=\frac{\cos(x)\cos(y)}{\left(r^2+x^2y^2+1)(1+e^x+e^y+e^{x+y}\right)},\,r^2=x^2+y^2 \end{align*} \] Aesthetically pleasing, probably of no use.
anonymous
  • anonymous
Alright, here's my first hint: Notice that \(\dfrac{\cos x}{x^2+1}\) is even.
anonymous
  • anonymous
Hint (1.a): \(y=-x\)
ganeshie8
  • ganeshie8
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 \[2I =\int_{-\infty}^\infty \frac{\cos x}{x^2+1}\,dx \]
ganeshie8
  • ganeshie8
since the integrand is even, \[I =\int_{0}^\infty \frac{\cos x}{x^2+1}\,dx \]
ganeshie8
  • ganeshie8
please do not give further hints, I think i got this :)
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
we may try DUT \[F(y) = \int_{0}^\infty \frac{\cos(xy)}{x^2+1}\,dx\] \[\implies F''(y) = -\int_{0}^\infty \frac{x^2\cos(xy)}{x^2+1}\,dx \] \[\implies -F''(y) + F(y) = \int_{0}^\infty \cos(xy)\,dx \]
ganeshie8
  • ganeshie8
scratch that, doesn't converge
anonymous
  • anonymous
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 \(t=\tan x\))
anonymous
  • anonymous
@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.
ganeshie8
  • ganeshie8
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
IrishBoy123
  • IrishBoy123
|dw:1442566006271:dw| do you not have to subtract them as the integration order needs to be reversed with the sub?
ganeshie8
  • ganeshie8
differential also changes sign so that fixes the bounds..
IrishBoy123
  • IrishBoy123
duh :-)
thomas5267
  • thomas5267
The answer is \(\dfrac{\pi}{e}\)? Read something extremely similar to this on Wikipedia and want to confirm the answer before trying.
thomas5267
  • thomas5267
Did I missed by a factor of 2?
anonymous
  • anonymous
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 :)
thomas5267
  • thomas5267
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.
thomas5267
  • thomas5267
Read at your own discretion!
anonymous
  • anonymous
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.
IrishBoy123
  • IrishBoy123
dark magic

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