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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've got some hints in mind, but I'm debating whether or not they give too much away :)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0I don't know but I kind of feel like it has something to do with Gaussian integral.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It'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
 one year ago
Best ResponseYou've already chosen the best response.0Convert to polar coordinate analogous to how you would evaluate Gaussian integral?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(x)=\frac{\cos x}{(x^2+1)(1+e^x)}\\ \begin{align*} f(x)f(y)&=\frac{\cos x}{(x^2+1)(1+e^x)}\frac{\cos y}{(y^2+1)(1+e^y)}\\ &=\frac{\cos(x)\cos(y)}{\left(r^2+x^2y^2+1)(1+e^x+e^y+e^{x+y}\right)},\,r^2=x^2+y^2 \end{align*} \] Aesthetically pleasing, probably of no use.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, here's my first hint: Notice that \(\dfrac{\cos x}{x^2+1}\) is even.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hint (1.a): \(y=x\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Picking 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
 one year ago
Best ResponseYou've already chosen the best response.1since the integrand is even, \[I =\int_{0}^\infty \frac{\cos x}{x^2+1}\,dx \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1please do not give further hints, I think i got this :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.1scratch that, doesn't converge

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oooh, 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442566006271:dw do you not have to subtract them as the integration order needs to be reversed with the sub?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1differential also changes sign so that fixes the bounds..

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0The answer is \(\dfrac{\pi}{e}\)? Read something extremely similar to this on Wikipedia and want to confirm the answer before trying.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Did I missed by a factor of 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, 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
 one year ago
Best ResponseYou've already chosen the best response.0Wikipedia: 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
 one year ago
Best ResponseYou've already chosen the best response.0Read at your own discretion!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, 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: \[\frac{1}{(x^2+1)(x^2+s^2)}=\frac{1}{s^21}\left(\frac{1}{x^2+1}\frac{1}{x^2+s^2}\right)\] And now we can integrate: \[\begin{align*}\mathcal{L}_s\{ I(a)\}&=\frac{1}{s^21}\int_0^\infty \left(\frac{1}{x^2+1}\frac{1}{x^2+s^2}\right)\,dx\\[1ex] &=\frac{1}{s^21}\left(\frac{\pi}{2}\frac{\pi}{2s}\right)\\[1ex] &=\frac{\pi}{2}\left(\frac{1}{s^21}\frac{1}{s(s^21)}\right)\\[1ex] &=\frac{\pi}{2}\left(\frac{1}{s^21}+\frac{1}{s}\frac{s}{s^21}\right) \end{align*}\] And finally, take the inverse transform. \[\begin{align*}\mathcal{L}_s\{ I(a)\}&=\frac{\pi}{2}\left(\frac{1}{s^21}+\frac{1}{s}\frac{s}{s^21}\right)\\[1ex] I(a)&=\mathcal{L}_a^{1}\{\cdots\}\\[1ex] &=\frac{\pi}{2}\left(\sinh a+1\cosh a \right) \end{align*}\] Let \(a=1\) and we're done.
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