Show that the limit \[ \lim_{t\to0^+}\left( \int_0^1\frac{dx}{(x^4+t^4)^{1/4}}+\log t \right) \] exists and is finite.
Solution
1. For \(t>0\), putting \(R=1/t\) gives\[ \int_0^1\frac{dx}{(x^4+t^4)^{1/4}}+\log t = \int_0^R\frac{du}{(1+u^4)^{1/4}}-\log R. \]
Make the substitution \[ x=tu. \] Then \[ dx=t\,du \] and \[ (x^4+t^4)^{1/4} = t(1+u^4)^{1/4}. \] Therefore \[ \int_0^1\frac{dx}{(x^4+t^4)^{1/4}} = \int_0^{1/t}\frac{du}{(1+u^4)^{1/4}}. \] Since \(\log t=-\log(1/t)\), the displayed identity follows.
2. For \(u\geq1\),\[ \left| \frac1{(1+u^4)^{1/4}}-\frac1u \right| \leq \frac1{4u^5}. \]
For \(u\geq1\), \[ \frac1{(1+u^4)^{1/4}} = \frac1u(1+u^{-4})^{-1/4}. \] Let \[ \psi(s)=(1+s)^{-1/4} \] for \(s\in[0,1]\). Since \[ \abs{\psi'(s)} = \frac14(1+s)^{-5/4} \leq \frac14, \] the mean value theorem gives \[ \abs{\psi(s)-1}\leq\frac{s}{4}. \] Taking \(s=u^{-4}\) and multiplying by \(1/u\) gives the claimed estimate.
3. The improper integral\[ \int_1^\infty \left( \frac1{(1+u^4)^{1/4}}-\frac1u \right)\,du \]converges absolutely.
By step 2, the absolute value of the integrand is bounded on \([1,\infty)\) by \[ \frac1{4u^5}. \] Since \[ \int_1^\infty\frac{du}{u^5}<\infty, \] the comparison test gives absolute convergence.
4. For every \(R\geq1\),\[ \int_0^R\frac{du}{(1+u^4)^{1/4}}-\log R = \int_0^1\frac{du}{(1+u^4)^{1/4}} + \int_1^R \left( \frac1{(1+u^4)^{1/4}}-\frac1u \right)\,du. \]
Split the first integral at \(1\) and use \[ \log R=\int_1^R\frac{du}{u}. \] Combining the two integrals over \([1,R]\) gives the displayed identity.
5. The limit in the problem exists and is finite.
3. Hence the required limit exists and is finite.
6. Q.E.D.
Step 5 proves the assertion.