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

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- anonymous

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

- jamiebookeater

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- anonymous

I've got some hints in mind, but I'm debating whether or not they give too much away :)

- myininaya

*

- thomas5267

I don't know but I kind of feel like it has something to do with Gaussian integral.

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## More answers

- 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

Convert to polar coordinate analogous to how you would evaluate Gaussian integral?

- 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

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

- anonymous

Hint (1.a): \(y=-x\)

- 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

since the integrand is even,
\[I =\int_{0}^\infty \frac{\cos x}{x^2+1}\,dx \]

- ganeshie8

please do not give further hints, I think i got this :)

- 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

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

scratch that, doesn't converge

- 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

@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

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

|dw:1442566006271:dw|
do you not have to subtract them as the integration order needs to be reversed with the sub?

- ganeshie8

differential also changes sign
so that fixes the bounds..

- IrishBoy123

duh
:-)

- thomas5267

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

- thomas5267

Did I missed by a factor of 2?

- 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

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

Read at your own discretion!

- 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

dark magic

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