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Is there a simple formula that would produce the regularized value for the most common divergent integrals?

I know, there is a formula for Cesaro integration, but it is applicable only to Cesaro-summable integrals. Other formulas for regularization essentially convert the integral into a series and regularize that series.

What I am looking for is an integral analog of Faulhaber's formula for Ramanujan's summation:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$

Or even its more universal variant:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)=-\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Yes, it is also not universally applicable, but it is universal enough. Is there an analog of this formula, but for integrals?

UPDATE

I came to the following formula: $$\operatorname{reg}\int_0^\infty f(x)\,dx=\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{1-e^{-2 \pi t}} \, dt-\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Unfortunately, the first term tends not to converge. The formula works though for integral $\int_0^\infty e^x dx$ (giving $-1$ as expected).

Is there a simple formula that would produce the regularized value for the most common divergent integrals?

I know, there is a formula for Cesaro integration, but it is applicable only to Cesaro-summable integrals. Other formulas for regularization essentially convert the integral into a series and regularize that series.

What I am looking for is an integral analog of Faulhaber's formula for Ramanujan's summation:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$

Or even its more universal variant:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)=\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Yes, it is also not universally applicable, but it is universal enough. Is there an analog of this formula, but for integrals?

UPDATE

I came to the following formula: $$\operatorname{reg}\int_0^\infty f(x)\,dx=\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{1-e^{-2 \pi t}} \, dt-\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Unfortunately, the first term tends not to converge. The formula works though for integral $\int_0^\infty e^x dx$ (giving $-1$ as expected).

Is there a simple formula that would produce the regularized value for the most common divergent integrals?

I know, there is a formula for Cesaro integration, but it is applicable only to Cesaro-summable integrals. Other formulas for regularization essentially convert the integral into a series and regularize that series.

What I am looking for is an integral analog of Faulhaber's formula for Ramanujan's summation:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$

Or even its more universal variant:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)=-\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Yes, it is also not universally applicable, but it is universal enough. Is there an analog of this formula, but for integrals?

Is there a simple formula that would produce the regularized value for the most common divergent integrals?

I know, there is a formula for Cesaro integration, but it is applicable only to Cesaro-summable integrals. Other formulas for regularization essentially convert the integral into a series and regularize that series.

What I am looking for is an integral analog of Faulhaber's formula for Ramanujan's summation:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$

Or even its more universal variant:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)=-\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Yes, it is also not universally applicable, but it is universal enough. Is there an analog of this formula, but for integrals?

UPDATE

I came to the following formula: $$\operatorname{reg}\int_0^\infty f(x)\,dx=\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{1-e^{-2 \pi t}} \, dt-\frac{i}{2} \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Unfortunately, the first term tends not to converge. The formula works though for integral $\int_0^\infty e^x dx$ (giving $-1$ as expected).

Is there a simple formula that would produce the regularized value for the most common divergent integrals?

I know, there is a formula for Cesaro integration, but it is applicable only to Cesaro-summable integrals. Other formulas for regularization essentially convert the integral into a series and regularize that series.

What I am looking for is an integral analog of Faulhaber's formula for Ramanujan's summation:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$

Or even its more universal variant:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)=-\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Yes, it is also not universally applicable, but it is universal enough.

Is there a simple formula that would produce the regularized value for the most common divergent integrals?

I know, there is a formula for Cesaro integration, but it is applicable only to Cesaro-summable integrals. Other formulas for regularization essentially convert the integral into a series and regularize that series.

What I am looking for is an integral analog of Faulhaber's formula for Ramanujan's summation:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n $$

Or even its more universal variant:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)=-\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Yes, it is also not universally applicable, but it is universal enough. Is there an analog of this formula, but for integrals?