this sum is said to be equal to $0$: $\int_{\mathbb{R}}\int_{-\infty}^x \frac{h(x)h(t)}{t-x}dt dx +\int_{\mathbb{R}}\int_{x}^{\infty} \frac{h(x)h(t)}{t-x}dt dx$ and I'm trying to understand why.
As reason I got said "because of reflection" but I still don't see it.
So I tried something:
$\int_{\mathbb{R}}\int_{-\infty}^x \frac{h(x)h(t)}{t-x}dt dx +\int_{\mathbb{R}}\int_{x}^{\infty} \frac{h(x)h(t)}{t-x}dt dx$
$=\int_{\mathbb{R}}\int_{-\infty}^x \frac{h(x)h(t)}{t-x}dt dx -\int_{\mathbb{R}}\int_{x}^{\infty} \frac{h(x)h(t)}{x-t}dt dx$
where I changed the sign of the second integral and therefore, I swapped $x$ and $t$ in the denominator.
$=\int_{\mathbb{R}}\int_{-\infty}^x \frac{h(x)h(t)}{t-x}dt dx +\int_{\mathbb{R}}\int_{\infty}^{x} \frac{h(x)h(t)}{x-t}dt dx$
where I swapped the integrationbounds of the second integral and therefore, the sign changes again.
$=\int_{\mathbb{R}}\int_{-\infty}^x \frac{h(x)h(t)}{t-x}dt dx -\int_{\mathbb{R}}\int_{\infty}^{x} \frac{h(x)h(t)}{t-x}dt dx$
where I swapped $t$ and $x$ in the denominator of the second integral back again and therefore, the sign of the second integral is negative again.
Did I do a mistake? I don't see why this should be equal to $0$, since in the second integral we have $\infty$ and $x$ as integrationbounds and in the first $-\infty$ and $x$ or should it also be $-\infty$ in the second integral?
Really thankful for any help!!
If the integrand is Lebesgue integrable then you can write the sum also as $$ \int_{\Bbb R ^2}\frac{h(x)h(t)}{t-x}\,\mathrm d (x,t)=\int_{x>t}\frac{h(x)h(t)}{t-x}\,\mathrm d (x,t)+\int_{x<t}\frac{h(x)h(t)}{t-x}\,\mathrm d (x,t)\\ =\int_{x<t}\frac{h(x)h(t)}{|t-x|}\,\mathrm d (x,t)+\int_{x>t}\frac{h(x)h(t)}{-|t-x|}\,\mathrm d (x,t)\\ =\int_{x<t}\frac{h(x)h(t)}{|t-x|}\,\mathrm d (x,t)-\int_{x>t}\frac{h(x)h(t)}{|t-x|}\,\mathrm d (x,t) $$ Now note that the last two integrals are exactly the same integral because $x$ and $t$ are tags that can be interchanged, so the final result is zero.