I would like to evaluate: ∫∞01√x(1+ex)dx Given that I can't find ∫1√x(1+ex)dx, a substitution is needed: I tried u=√x(1+ex) and u=1√x(1+ex) but I could not get rid of x in the new integral.... Do you have ideas of substitution?
Answer
∫∞01√x(1+ex)dx=2∫∞01√1+ex2dx=2∫∞0(1+e−x2)−1/2e−x2/2dx=2∫∞0∞∑k=0(−14)k(2kk)e(2k+1)x2/2dx=∞∑k=0(−14)k(2kk)√2π2k+1
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