Relations between definite integrals not having a known closed form Are there any known cases, when there are two (or more) definite integrals, none of them having any known closed-form expression on its own, but there is still a non-trivial$^\dagger$ elementary relation connecting them (e.g. one integral is the natural logarithm of the other)?

$^\dagger$ non-trivial means we want to exclude non-interesing cases (especially, linear combinations) like one integral being twice larger then the other because its integrand is twice as large.
 A: It seems natural to consider the gamma function and the log gamma function.  Recall that $$\Gamma(t) = \int_{0}^{\infty} x^{t-1} e^{-x} dx$$ for complex numbers with a positive real part.  Binet's second log gamma formula (see http://mathworld.wolfram.com/BinetsLogGammaFormulas.html) is $$\ln \Gamma(z) = \left( z - \frac{1}{2} \right) \ln z - z + \frac{1}{2} \ln (2 \pi) + 2 \int_{0}^{\infty} \frac{\tan^{-1}\left(\frac{t}{z}\right)}{e^{2 \pi t} - 1} dt$$ and we thus have that: $$\ln\left( \int_{0}^{\infty} x^{t-1} e^{-x} dx\right) = \left( z - \frac{1}{2} \right) \ln z - z + \frac{1}{2} \ln (2 \pi) + 2 \int_{0}^{\infty} \frac{\tan^{-1}\left(\frac{t}{z}\right)}{e^{2 \pi t} - 1} dt.$$ Certain values of the gamma function such as $\Gamma\left(\frac{1}{3}\right)$ have no closed form, so the above definite integrals have no known closed form for certain values of $t$.
A: This is the most fitting nonlinear identity I found so far. There's still a closed form here somewhere, but without knowing the details, it's not at all obvious.
Here's the identity:

$$\int_{-\infty}^\infty \frac{e^{a x} \cos x}{(e^x+1)^{2a}}dx= \\ \frac{2 \sqrt{\pi}}{4^a} \frac{\Gamma(a)}{\Gamma(a+\frac12 )} \exp \left(-2 \int_0^\infty \frac{e^{(1-a) x} (1-\cos x)}{x (e^x-1)}dx \right), \quad a>0$$

If anyone is interested, I can provide the proof.
I'm sure some people will guess the closed form for the left hand side right away.
As a simple case, and also a hint, it follows for $a=1$ (Mathematica can't take even this integral):
$$\int_0^\infty \frac{1-\cos x}{x (e^x-1)}dx=\frac{1}{2} \log \frac{\sinh \pi}{\pi}$$
But the closed form is definitely not as simple for the general $a \in \mathbb{R}^+$.
A: Using the infinite product relation (see this question):
$$\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$$
And provided that the two following integrals exist for some function $f(x)$ and some limits $a,b$:
$$\int_a^b f(x) \ln \left(1-x^n+x^{2n}\right)dx$$
$$\int_a^b f(x) \ln \left(1+x^{2n-1}+x^{4n-2}\right)dx$$
We get the following relation:

$$\sum_{n=1}^{\infty} \int_a^b f(x) \ln \left(1-x^n+x^{2n}\right)dx=-\sum_{n=1}^{\infty} \int_a^b f(x) \ln \left(1+x^{2n-1}+x^{4n-2}\right)dx$$

By definition no closed form exists for these integrals in general. 
This is technically a linear combination, but I wouldn't consider this example trivial.
Using some other infinite product relations we can find other cases for $\log$ integrals.

I would also advise looking for multiple integrals (generalizing Poisson's trick):
$$I=\int_a^b g(x)~dx$$
$$I^k=\int_a^b \dots \int_a^b g(x_1) \dots g(x_k)~dx_1 \dots dx_k$$
If we can transform the latter expression to an appropriate coordinate system, we can theoretically reduce the number of integrations (especially if $a=-\infty$, $b=+\infty$) and get a number of non-trivial non-linear relations, even adding different powers and getting some Taylor series).
The method can also be used for different integrals:
$$I \cdot J=\int_a^b \int_\alpha^\beta g(x) f(y)~ dx~ dy$$

I think a linear example is better than no example, so I'll add this:

$$\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x}}}}+\int_1^{3+2\sqrt{2}} \frac{(v^2+1)~dv}{v\sqrt{v-1}\sqrt{v^3+v^2+7v-1}}=2\sqrt{1+\sqrt{2}}$$

As far as I know there is no closed form for both integrals. The result is obtained by simple substitution.
A: One such relation was posted in this question. Taking different combinations of hyperbolic functions leads to other relations of this kind.
Also, proceeding along the same lines one can obtain another quadratic relation
$$
\left(\int\limits_0^\infty\frac{\cos\alpha x^2}{x^2+1}dx\right)^2+\left(\int\limits_0^\infty\frac{\sin\alpha x^2}{x^2+1}dx\right)^2={\pi\int\limits_0^\infty\frac{e^{-2\alpha x}}{x^2+4}\left(\cos\alpha x^2+\frac2x \sin\alpha x^2\right)dx}
$$
