Find F'(t), where: F(t) = double integral over

Find $F'(t)$, where: $$ F(t) = \iint_{[0,t]\times[0,t]} e^{tx/y^2}\,dx\,dy, \qquad t>0. $$

Step 1 · Examine the integrand near $y=0$

Fix $t>0$ and any $x\in(0,t]$. As $y\to0^+$, the exponent $\dfrac{tx}{y^2}\to+\infty$ (since $t>0$ and $x>0$), and therefore:

$$ e^{tx/y^2} \longrightarrow +\infty \quad\text{as } y\to0^+, $$

and it does so faster than any power of $\dfrac1y$ — the exponential blow-up dominates any algebraic decay one might hope to offset it with.

Step 2 · Show the inner integral diverges

For fixed $x\in(0,t]$, consider the inner integral with respect to $y$:

$$ \int_0^t e^{tx/y^2}\,dy. $$

Since $e^{tx/y^2}>1$ for all $y\in(0,t]$ (the exponent is positive), and $e^{tx/y^2}\to+\infty$ non-integrably as $y\to0^+$ (it is not dominated by any integrable function near $y=0$; indeed for any $M$, $e^{tx/y^2}>M$ once $y<\sqrt{tx/\ln M}$, an interval of positive length), the integral diverges:

$$ \int_0^t e^{tx/y^2}\,dy = +\infty \qquad \text{for every fixed } x\in(0,t]. $$

Step 3 · Conclude that $F(t)$ is not finite

Since the inner integral is already $+\infty$ for (almost) every $x\in[0,t]$, the double integral diverges as well:

$$ F(t) = \int_0^t\left(\int_0^t e^{tx/y^2}\,dy\right)dx = +\infty \qquad \text{for every } t>0. $$

Numerically, truncating the $y$-integration away from $0$ confirms explosive, non-convergent growth: cutting off at $y\geqslant0.1$ already gives a partial value of order $10^{38}$ (for $t=1$), and the partial values grow without bound as the cutoff shrinks toward $0$.

Additional Notes

• Because $F(t)\equiv+\infty$, the function $F$ is not real-valued and $F'(t)$ does not exist; Leibniz's rule for differentiating under the integral sign cannot be applied here, since that rule requires the integral to converge (indeed, to converge locally uniformly in $t$) in the first place.
• The divergence traces to a single structural feature: the exponent $tx/y^2$ is unbounded as $y\to0$ for any fixed $x>0$, and no power of $y$ in the exponent's denominator higher than $y^1$ can be tolerated for the resulting exponential blow-up to remain integrable near the origin. Had the exponent instead been, say, $-tx/y^2$ or $tx\cdot y^2$, the integral would converge and Leibniz's rule would apply directly.
• This is a useful cautionary example: formally differentiating under the integral sign without first confirming convergence of $F(t)$ itself can produce a symbolic "answer" that refers to a function which is not actually well-defined.