Q15·Section A (Section A)·15 of 60Convergence of two improper integralsConsider the improper integrals I1=∫1∞tsintet dtandI2=∫1∞1tln(1+1t) dt.I_1=\int_1^\infty\frac{t\sin t}{e^t}\,dt\quad\text{and}\quad I_2=\int_1^\infty\frac1{\sqrt{t}}\ln\Bigl(1+\frac1t\Bigr)\,dt.I1=∫1∞ettsintdtandI2=∫1∞t1ln(1+t1)dt. Then OptionsAI1I_1I1 converges but I2I_2I2 does NOT convergeBI1I_1I1 does NOT converge but I2I_2I2 convergesCboth I1I_1I1 and I2I_2I2 convergeDneither I1I_1I1 nor I2I_2I2 convergesPick an option to check your answer.Show answerShow solution Report a problem with this question← Q14All questionsQ16 →Keyboard: A–D answer · S solution · ←/→ previous/next