# From Sophie Germain's Identity to difficult integrals: the integral $\int_0^1\int_0^1\cos\left((x+y)^2+y^2\right)dxdy$

Yesterday I was interested in the so-called Sophie Germain's Identity, see this section of Wikipedia. I was interested in it about composite numbers. After I was thinking about different consequences of this identity. Playing I prosoposed calculate the easy $$\int_0^1\int_0^1\frac{x^4+4y^4}{x^2-2xy+2y^2}dxdy,$$ and after I was interested in integrals $$\int_0^1\int_0^1 f\left(\frac{x^4+4y^4}{x^2-2xy+2y^2}\right)dxdy,$$ for different functions than $f(z)=z$. For example $f(z)=\log z$, that can be solved in closed-form.

A different example was $f(z)=\cos z$, using a CAS I believe that it can not solve in closed-form.

Question. Calculate in terms of series, special functions (integral functions) or well-known constants $$\int_0^1\int_0^1\cos\left((x+y)^2+y^2\right)dxdy.\tag{1}$$ Many thanks.

I am thinking how get an approximation of $(1)$ using calculus, thus avoid provide me hints to this purpose.

• The upper limit $x=1, y=1$ meant nothing for the cosine function, so I would only expect a closed form in terms of very contrived special functions. – pisco Oct 31 '17 at 10:56
• I accept your critic, since I've posted this question, feel free if you want ask or study different versions. Many thanks for your attention @pisco125 – user243301 Oct 31 '17 at 11:07

Notation: $C$ and $S$ are Fresnel integral.

let calculate first integrate thus we can write:

$\int cos((x+y)^2+y)dx$

using $u=x+y \to dx=du$ we can write it like :

$\int(cos(u+y)du$ expand trigonometric function: $\int(cos(y^2)cos(u^2)-sin(y^2)sin(u^2))du=cos(y^2)\int cos(u^2)du-sin(y^2)\int sin(u^2)du$

calculate first and second integrals with $v={u\sqrt2 \over \sqrt \pi}$ .

and finally with a little calculate :

## $- \pi ({sin(y^2)S({\sqrt 2 (x+y) \over \sqrt \pi})-cos(y^2)C({\sqrt 2 (x+y) \over \sqrt \pi}))}\over \sqrt 2$

for the first integral .

to calculate the second one we split them and calculate them sepratly so we write : \begin{align} \int sin(y^2)S({\sqrt 2 (x+y) \over \sqrt \pi})dy=sin(y^2)((x+y)S((x+y) \sqrt { 2 \over \pi } )+{cos((x+y)^2) \over \sqrt {2\pi}}) \end{align}

and exacly like first one you can write it for second one : \begin{align} \int cos(y^2)C({\sqrt 2 (x+y) \over \sqrt \pi})dy=cos(y^2)((x+y)C((x+y) \sqrt { 2 \over \pi } )-{sin((x+y)^2) \over \sqrt {2\pi}}) \end{align}

and finally you just need to put all of this together and calculate them.

• Many thanks I am going to study your answer this afternoon. – user243301 Oct 31 '17 at 12:24
• you're welcome @user243301 – Dexpectra Oct 31 '17 at 12:28
• Go ahead and precede upright functions like $\cos$ with a backslash \cos, and you can use \left and \right followed by parentheses, brackets, etc. for them to automatically expand to the size of the enclosed content. – gen-ℤ ready to perish Oct 31 '17 at 12:42
• Check the edits I made to this answer as well as the MathJax tutorial – gen-ℤ ready to perish Oct 31 '17 at 12:49