# What is $\int_0^{\pi} \frac{e^{\sin x}\cos(x)}{1+e^{\tan x}} \, dx$?

I read this question. The integral has a special property, so it might possibly be evaluable? No one tried evaluating it, so I created this. Not very often I ask question like this, but here it is.

So what is,

$$\int_0^{\pi} \frac{e^{\sin x}\cos(x)}{1+e^{\tan x}} \, dx$$

A list of Naïve Ideas

Sub $u=\sin(x)$ allows you to cancel the $\cos$.

Half angle substitution

List the obscure special function it equals.

Try contour integration ;) (just being stupid here)

(Disclaimer: I have no reason to believe this integral has closed form beyond the property mentioned in the link)

• Is not necessary the modulus in the integral! – Euler88 ... Aug 1 '15 at 20:42
• What is the reason for the absolute value, since sine is already positive on that interval? – John Molokach Aug 1 '15 at 20:42
• @JohnMolokach If you look at the link, you can see he just copy-pasted the integral from that question and changed the bounds. In that question, the absolute value is necessary to enable a trick to calculate the integral - here, it isn't. – izœc Aug 1 '15 at 20:46
• It may be of interest to note that, for $\phi(x) = e^{|\sin x|} \, \cos x$, $$\int_{0}^{\pi} \frac{\phi(x) \, dx}{1 + e^{\tan x}} \, dx = - \frac{1}{2} \, \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2}\right) \, \phi(x) \, dx.$$ – Leucippus Aug 1 '15 at 20:59
• @Zach466920 The function does not have any obvious symmetry on $[0,\pi]$ and WA returns a numerical value only, not any recognizable form. Have you tried a series solution approach or other function theoretic way forward? – Mark Viola Aug 1 '15 at 22:10

As the integral was first defined, with $\phi(x) = e^{|\sin x|} \, \cos x$, an identity was presented as, in corrected form, \begin{align} \int_{0}^{\pi} \frac{\phi(x) \, dx}{1 + e^{\tan x}} = - \frac{1}{2} \, \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2} \right) \, \phi(x) \, dx. \end{align} Wolfram Alpha provides a numerical value of $- .727486$ for the integral on the left and \begin{align} \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2} \right) \, \phi(x) \, dx = 1.45497. \end{align} In either case the numerical values lead to the speculation that \begin{align} \int_{0}^{\pi} \frac{\phi(x) \, dx}{1 + e^{\tan x}} = - \frac{1}{2} \, \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2} \right) \, \phi(x) \, dx \approx - \frac{672}{855}. \end{align} Further work can be done to evaluate the integral.