@@cipherunity All i did was to taylor expand the log(x+1), integrating which we get an anwer that can be written as diff of 2 alternating inf series, one of which will be log2 and other will be diff of 2 diagamma. My answer comes (sqrt(2)/3)[log4-digamma(5/4)+digamma(7/4)]
@@cipherunity 2nd alternating inf series would be sum[(-1)^n+1/(n+ 3/2)] that can be written as difference of 2 inf series sum(1/(5/2 + 2n)) - sum(1/(7/2 + 2n) which equates to (1/2)(digamma(5/4) - digamma(7/4))
Excellent
Hi sir any idea [math] \int_0^\infty \frac{\tan x \sin x d x}{\tan x \sin x+(x \cot x+1)^2}[/math]?
I shall look into it
Good question, i got exact solution but using diagamma function
If you provide your solution we can make a new video in your name.
@@cipherunity All i did was to taylor expand the log(x+1), integrating which we get an anwer that can be written as diff of 2 alternating inf series, one of which will be log2 and other will be diff of 2 diagamma. My answer comes (sqrt(2)/3)[log4-digamma(5/4)+digamma(7/4)]
@@divyaanshu123 I shall see to it
@@divyaanshu123 Every step is clear. Except the digamma part. Can you put some more light on it.
@@cipherunity 2nd alternating inf series would be sum[(-1)^n+1/(n+ 3/2)] that can be written as difference of 2 inf series sum(1/(5/2 + 2n)) - sum(1/(7/2 + 2n) which equates to (1/2)(digamma(5/4) - digamma(7/4))