$$ \int _ { 0 } ^ { \infty } E d m \quad , \quad E = \sqrt { p ^ { 2 } \cdot c ^ { 2 } + m ^ { 2 } \cdot c ^ { 2 } } $$

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UpStudy AI · Accepted Answer

Certainly! Let's evaluate the integral you provided:

$$
\int_{0}^{\infty} E \, dm \quad \text{where} \quad E = \sqrt{p^2 c^2 + m^2 c^2}
$$

### Simplifying the Expression

First, factor out $$ c $$ from the expression for $$ E $$:

$$
E = \sqrt{p^2 c^2 + m^2 c^2} = c \sqrt{p^2 + m^2}
$$

Therefore, the integral becomes:

$$
\int_{0}^{\infty} E \, dm = c \int_{0}^{\infty} \sqrt{p^2 + m^2} \, dm
$$

### Evaluating the Integral

The integral we need to evaluate is:

$$
\int_{0}^{\infty} \sqrt{m^2 + p^2} \, dm
$$

This is a standard integral in calculus. The antiderivative of $$ \sqrt{m^2 + a^2} $$ with respect to $$ m $$ is:

$$
\frac{m}{2} \sqrt{m^2 + a^2} + \frac{a^2}{2} \ln\left(m + \sqrt{m^2 + a^2}\right) + C
$$

Where $$ a = p $$ in our case.

Applying the limits from $$ 0 $$ to $$ \infty $$:

$$
\left. \left( \frac{m}{2} \sqrt{m^2 + p^2} + \frac{p^2}{2} \ln\left(m + \sqrt{m^2 + p^2}\right) \right) \right|_{0}^{\infty}
$$

1. **As $$ m \to \infty $$:**
   - $$ \frac{m}{2} \sqrt{m^2 + p^2} \approx \frac{m}{2} \cdot m = \frac{m^2}{2} $$ → $$ \infty $$
   - $$ \frac{p^2}{2} \ln\left(m + \sqrt{m^2 + p^2}\right) \approx \frac{p^2}{2} \ln(2m) $$ → $$ \infty $$

2. **At $$ m = 0 $$:**
   - $$ \frac{0}{2} \sqrt{0 + p^2} = 0 $$
   - $$ \frac{p^2}{2} \ln\left(0 + \sqrt{0 + p^2}\right) = \frac{p^2}{2} \ln(p) $$

### Conclusion

As $$ m \to \infty $$, both terms in the antiderivative tend to infinity. Therefore, the definite integral diverges:

$$
\int_{0}^{\infty} \sqrt{m^2 + p^2} \, dm = \infty
$$

Multiplying by $$ c $$, the original integral also diverges:

$$
\int_{0}^{\infty} E \, dm = c \cdot \infty = \infty
$$

### Physical Interpretation

In a physical context, this divergence suggests that the integral doesn't converge to a finite value under the given conditions. This might imply that integrating energy $$ E $$ over all possible mass values $$ m $$ from $$ 0 $$ to $$ \infty $$ isn't physically meaningful without additional constraints or considerations.

If you have a specific physical scenario or additional constraints in mind, please provide more details, and I can offer a more tailored explanation or solution.
The integral $$ \int_{0}^{\infty} E \, dm $$ where $$ E = \sqrt{p^2 c^2 + m^2 c^2} $$ diverges to infinity.

Upstudy AI Solution

Answer

Solution

Simplifying the Expression

Evaluating the Integral

Conclusion

Physical Interpretation

Beyond the Answer

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