How do you integrate #int x^c e^(x^d)dx#, where #c>d#, using integration by parts?

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Answer 1 · 2016-10-04T08:27:17

Use the recurrence formula

#I_(d+k)+(k+1)/dI_k = 1/dx^(k+1)e^(x^d)#

with initial #k = c-d# and

#I_(d-1)= 1/de^(x^d)#

We use #n, m# instead of #c,d#,
.

#d/(dx)(x^n e^(x^m)) = nx^(n-1)e^m+ m x^(n+m-1) e^(x^m)#

Calling now

#I_n = int x^n e^(x^m)dx# we have the recurrence relationship

#mI_(n+m-1)+nI_(n-1)=x^n e^(x^m)#

with

#I_(m-1)=int x^(m-1)e^(x^m)dx = 1/me^(x^m)# or calling #k = n-1#

#mI_(m+k)+(k+1)I_k = x^(k+1)e^(x^m)#

Finally

#I_(m+k)+(k+1)/mI_k = 1/mx^(k+1)e^(x^m)#

Example. #n=5, m=3# so #k=2# and we want

#I_(m+2)= I_5 = int x^5 e^(x^3)dx#

we know that #I_(m-1) = I_2# and

#I_2=1/3e^(x^3)# so

#I_(m+2)+(2+1)/mI_2=1/mx^3e^(x^3)# so

#I_5 +3/3I_2=1/3x^3e^(x^3)# or

#I_5 = 1/3x^3 e^(x^3)-1/3e^(x^3)=1/3(x^3-1)e^(x^3)#