.

You have to divide by

to get

, thus

. All you have to do now is see that

.

In order to make this last step easier, you can notice the following: if we let

,

,

,

, then

,

and

, and we have the following beautiful identity (for any

):

,

so that

. qed.

About the previous identity with a,b,c,d (dating back to Euler): it can be seen as an expression of the formula

where

,

. Of course this identity is not absolutely necessary here, you can just expand and see many terms simplify. There is a reason why this identity is involved: the integral can be written

where

,

. But the justification of the computation is more delicate if we do it this way.

As a conclusion,

.

(NB: there is a minus sign in front of the exponent of the answer; it wasn't in your post)

Feel free to ask for details if something's unclear.