We will analyze the improper integral:

$\int_{0}^{3} e^{-3/t^4} t^5 dt$

Step (a): Identifying Limits of Integration

We split the given integral at an intermediate point $b$ (typically 1), setting:

$\int_0^3 e^{-3/t^4} t^5 dt = \int_0^b e^{-3/t^4} t^5 dt + \int_b^3 e^{-3/t^4} t^5 dt$

From the given notation:

• $a = 0^+$ (since the lower limit approaches 0)
• $b = 1$ (a convenient midpoint)
• $c = 3$ (upper limit)

Step (b): Substituting a New Variable

We use the substitution:

$u = \frac{3}{t^4}$

Then, differentiating:

$du = -\frac{12}{t^5} dt \quad \Rightarrow \quad dt = -\frac{t^5}{12} du$

Since $t = (3/u)^{1/4}$, we get the transformed limits:

• $t = b = 1 \Rightarrow u = 3/1^4 = 3$
• $t = c = 3 \Rightarrow u = 3/3^4 = 3/81 = 1/27$

Thus, we transform the integral.

Step (c): Transforming the Integral

Rewriting in terms of $u$:

$\int_1^3 e^{-3/t^4} t^5 dt = \int_{1/27}^{3} e^{-u} \left(-\frac{(3/u)^{5/4}}{12} du\right)$

Rewriting the function:

• $B = 1/27$
• $C = 3$
• $f(u) = e^{-u} \left(-\frac{(3/u)^{5/4}}{12} \right)$

Step (d): Evaluating the Integral

The integral remains improper at $t = 0$, where the denominator in the exponent approaches infinity. This suggests divergence.

Step (e): Final Limit Calculation

$\lim_{T \to 0^+} \int_{T}^{3} e^{-3/t^4} t^5 dt$

Since the function grows too large near $t = 0$, the integral diverges.

Step (f): Conclusion

$\int_0^3 e^{-3/t^4} t^5 dt = \textbf{Diverges}$

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Would you like a deeper breakdown of any step? Here are some related questions:

1. How does improper integration work in general?
2. Why does the function $e^{-3/t^4}$ cause divergence?
3. How do we decide on substitution choices for integrals?
4. What happens if we change the exponent in $e^{-3/t^4}$?
5. How does this compare to other common improper integrals?

Tip: When dealing with improper integrals, always check the behavior at singularities like $t = 0$ and $t \to \infty$.