There is a well-known theorem that states that for $f$ continuous and $f,\hat f$ integrable, $$f(0)=\frac{1}{\pi}\lim_{T\to\infty}\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx.$$ So it should be that $$\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx=f(0)+\text{(Error term)}.$$ Are there known explicit formulas for estimating this error term? Answers with added assumptions (e.g, smooth, even, etc.) are welcome.

There is a well-known theorem that states that for $f$ continuous and $f,\hat f$ integrable, $$f(0)=\frac{1}{\pi}\lim_{T\to\infty}\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx.$$ So it should be that $$\frac{1}{\pi}\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx=f(0)+\text{(Error term)}.$$ Are there known explicit formulas for estimating this error term? Answers with added assumptions (e.g, smooth, even, etc.) are welcome.

There is a well-known theorem that states that for $f$ continuous and $f,\hat f$ integrable, $$f(0)=\lim_{T\to\infty}\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx.$$ So it should be that $$\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx=f(0)+\text{(Error term)}.$$ Are there known explicit formulas for estimating this error term?

There is a well-known theorem that states that for $f$ continuous and $f,\hat f$ integrable, $$f(0)=\frac{1}{\pi}\lim_{T\to\infty}\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx.$$ So it should be that $$\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx=f(0)+\text{(Error term)}.$$ Are there known explicit formulas for estimating this error term? Answers with added assumptions (e.g, smooth, even, etc.) are welcome.