# Seemingly hard integrals which are made easy via differentiation under the integral sign a.k.a Feynman Integration [closed]

I recently discovered Differentiation under the integral sign a.k.a Feynman Integration and I read an article which says it can be substituted for contour integration. Therefore, I am assuming this technique is, indeed, very powerful. I was looking for a list of integrals which are, seemingly, hard but are made easy via this technique.

Thanks a lot!

• One of them is math.stackexchange.com/questions/874029/… – Jack D'Aurizio Jul 25 '14 at 12:26
• Some easier examples of such integrals are $$\int^{\infty}_0\frac{\ln{x}}{1+x^4}dx$$ and $$\int^1_0x^a(\ln{x})^bdx$$ – SuperAbound Jul 25 '14 at 12:27
• Wikipedia gives a dozen of examples of different types. You don't find this sufficient? – Start wearing purple Jul 25 '14 at 12:28
• @O.L. I thought it would be nice to have a list on math.se with explanations provided by experienced users to look at and furthermore, for people new to the technique to look it's power and see how it applies to a wide range of integrals. – Jeel Shah Jul 25 '14 at 13:08
• You should certainly look at math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf for some nice insights by Keith Conrad – James S. Cook Jul 25 '14 at 13:20

Note that $$\int_0^1 x^\alpha\ dx=\frac1{\alpha+1},\qquad\text{for }\ \alpha>-1.\tag1$$ Differentiating $(1)$ $n$ times yields $$\int_0^1 \frac{\partial^n}{\partial\alpha^n}\left(x^\alpha\right)\ dx=\color{blue}{\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}}, \qquad\text{for }\ n=0,1,2,\ldots\tag2$$